What Is Physics?
One job of physics is to identify the different types of energy in the world,
especially those that are of common importance. One general type of energy is
potential energy U. Technically, potential energy is energy that can be associated
with the configuration (arrangement) of a system of objects that exert forces on
one another.
CHAPTER 8
Potential Energy and
Conservation of Energy
8-1 POTENTIAL ENERGY
After reading this module, you should be able to . . .
8.01 Distinguish a conservative force from a nonconservative
force.
8.02 For a particle moving between two points, identify that
the work done by a conservative force does not depend on
which path the particle takes.
8.03 Calculate the gravitational potential energy of a particle
(or, more properly, a particle–Earth system).
8.04 Calculate the elastic potential energy of a block–spring
system.
A force is a conservative force if the net work it does on
a particle moving around any closed path, from an initial
point and then back to that point, is zero. Equivalently, a force
is conservative if the net work it does on a particle moving
between two points does not depend on the path taken by
the particle. The gravitational force and the spring force are
conservative forces; the kinetic frictional force is a noncon-
servative force.
Potential energy is energy that is associated with the con-
figuration of a system in which a conservative force acts.
When the conservative force does work Won a particle
within the system, the change Uin the potential energy of
the system is
UW.
If the particle moves from point xito point xf, the change in
the potential energy of the system is
U
xf
xi
F(x)dx.
The potential energy associated with a system consisting of
Earth and a nearby particle is gravitational potential energy. If
the particle moves from height yito height yf, the change in the
gravitational potential energy of the particle–Earth system is
Umg(yfyi)mg y.
If the reference point of the particle is set as yi0and the
corresponding gravitational potential energy of the system is
set as Ui0, then the gravitational potential energy Uwhen
the particle is at any height yis
U(y)mgy.
Elastic potential energy is the energy associated with the
state of compression or extension of an elastic object. For a
spring that exerts a spring force Fkx when its free end
has displacement x, the elastic potential energy is
The reference configuration has the spring at its relaxed
length, at which x0and U0.
U(x)1
2kx2.
Key Ideas
Learning Objectives
177177
This is a pretty formal definition of something that is actually familiar to you.
An example might help better than the definition: A bungee-cord jumper plunges
from a staging platform (Fig. 8-1). The system of objects consists of Earth and the
jumper.The force between the objects is the gravitational force.The configuration
of the system changes (the separation between the jumper and Earth decreases
that is, of course, the thrill of the jump). We can account for the jumper’s motion
and increase in kinetic energy by defining a gravitational potential energy U. This
is the energy associated with the state of separation between two objects that at-
tract each other by the gravitational force, here the jumper and Earth.
When the jumper begins to stretch the bungee cord near the end of the
plunge, the system of objects consists of the cord and the jumper. The force
between the objects is an elastic (spring-like) force. The configuration of the sys-
tem changes (the cord stretches). We can account for the jumper’s decrease in
kinetic energy and the cord’s increase in length by defining an elastic potential
energy U.This is the energy associated with the state of compression or extension
of an elastic object, here the bungee cord.
Physics determines how the potential energy of a system can be calculated so
that energy might be stored or put to use. For example, before any particular
bungee-cord jumper takes the plunge, someone (probably a mechanical engi-
neer) must determine the correct cord to be used by calculating the gravitational
and elastic potential energies that can be expected.Then the jump is only thrilling
and not fatal.
Work and Potential Energy
In Chapter 7 we discussed the relation between work and a change in kinetic energy.
Here we discuss the relation between work and a change in potential energy.
Let us throw a tomato upward (Fig. 8-2).We already know that as the tomato
rises, the work Wgdone on the tomato by the gravitational force is negative
because the force transfers energy from the kinetic energy of the tomato.We can
now finish the story by saying that this energy is transferred by the gravitational
force to the gravitational potential energy of the tomatoEarth system.
The tomato slows, stops, and then begins to fall back down because of the
gravitational force. During the fall, the transfer is reversed:The work Wgdone on
the tomato by the gravitational force is now positivethat force transfers energy
from the gravitational potential energy of the tomatoEarth system to the
kinetic energy of the tomato.
For either rise or fall, the change Uin gravitational potential energy is
defined as being equal to the negative of the work done on the tomato by the
gravitational force. Using the general symbol Wfor work, we write this as
UW. (8-1)
178 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Figure 8-1 The kinetic energy of a bungee-
cord jumper increases during the free fall,
and then the cord begins to stretch, slow-
ing the jumper.
Rough Guides/Greg Roden/Getty Images, Inc.
Figure 8-2 A tomato is thrown upward. As it rises, the
gravitational force does negative work on it, decreasing
its kinetic energy. As the tomato descends, the
gravitational force does positive work on it, increasing
its kinetic energy.
Negative
w
ork done
by the
gravitational
force
Positive
work done
by the
gravitational
force
This equation also applies to a blockspring system, as in Fig. 8-3. If we
abruptly shove the block to send it moving rightward, the spring force acts leftward
and thus does negative work on the block, transferring energy from the kinetic
energy of the block to the elastic potential energy of the springblock system.The
block slows and eventually stops, and then begins to move leftward because the
spring force is still leftward. The transfer of energy is then reversedit is from
potential energy of the springblock system to kinetic energy of the block.
Conservative and Nonconservative Forces
Let us list the key elements of the two situations we just discussed:
1. The system consists of two or more objects.
2. Aforce acts between a particle-like object (tomato or block) in the system and
the rest of the system.
3. When the system configuration changes, the force does work (call it W1) on
the particle-like object, transferring energy between the kinetic energy Kof
the object and some other type of energy of the system.
4. When the configuration change is reversed, the force reverses the energy
transfer,doing work W2in the process.
In a situation in which W1W2is always true, the other type of energy is
a potential energy and the force is said to be a conservative force. As you might
suspect, the gravitational force and the spring force are both conservative (since
otherwise we could not have spoken of gravitational potential energy and elastic
potential energy, as we did previously).
A force that is not conservative is called a nonconservative force. The kinetic
frictional force and drag force are nonconservative. For an example, let us send
a block sliding across a floor that is not frictionless. During the sliding, a kinetic
frictional force from the floor slows the block by transferring energy from its
kinetic energy to a type of energy called thermal energy (which has to do with the
random motions of atoms and molecules). We know from experiment that this
energy transfer cannot be reversed (thermal energy cannot be transferred back
to kinetic energy of the block by the kinetic frictional force). Thus, although we
have a system (made up of the block and the floor), a force that acts between
parts of the system, and a transfer of energy by the force, the force is not conser-
vative.Therefore, thermal energy is not a potential energy.
When only conservative forces act on a particle-like object, we can greatly
simplify otherwise difficult problems involving motion of the object. Let’s next
develop a test for identifying conservative forces, which will provide one means
for simplifying such problems.
Path Independence of Conservative Forces
The primary test for determining whether a force is conservative or nonconserva-
tive is this: Let the force act on a particle that moves along any closed path, begin-
ning at some initial position and eventually returning to that position (so that the
particle makes a round trip beginning and ending at the initial position). The
force is conservative only if the total energy it transfers to and from the particle
during the round trip along this and any other closed path is zero. In other words:
179
8-1 POTENTIAL ENERGY
The net work done by a conservative force on a particle moving around any
closed path is zero.
Figure 8-3 A block, attached to a spring and
initially at rest at x0, is set in motion
toward the right. (a) As the block moves
rightward (as indicated by the arrow), the
spring force does negative work on it.
(b) Then, as the block moves back toward
x0, the spring force does positive work
on it.
(a)
(b)
0
x
0
x
We know from experiment that the gravitational force passes this closed-
path test. An example is the tossed tomato of Fig. 8-2. The tomato leaves the
launch point with speed v0and kinetic energy .The gravitational force acting
1
2mv0
2
Checkpoint 1
The figure shows three paths connecting points a
and b.A single force does the indicated work on
a particle moving along each path in the indicated
direction. On the basis of this information, is force
conservative?
F
:
F
:
on the tomato slows it, stops it, and then causes it to fall back down. When the
tomato returns to the launch point, it again has speed v0and kinetic energy
Thus, the gravitational force transfers as much energy from the tomato dur-
ing the ascent as it transfers to the tomato during the descent back to the launch
point. The net work done on the tomato by the gravitational force during the
round trip is zero.
An important result of the closed-path test is that:
1
2mv0
2.
180 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
b
a
1
2
(a)
b
a
1
2
(b)
The force is
conservative. Any
choice of path
between the points
gives the same
amount of work.
And a round trip
gives a total work
of zero.
Figure 8-4 (a) As a conservative force acts
on it, a particle can move from point ato
point balong either path 1 or path 2.
(b) The particle moves in a round trip,
from point ato point balong path 1 and
then back to point aalong path 2.
a
b
60
J
60 J
60 J
The work done by a conservative force on a particle moving between two points
does not depend on the path taken by the particle.
For example, suppose that a particle moves from point ato point bin Fig. 8-4a
along either path 1 or path 2. If only a conservative force acts on the particle,then
the work done on the particle is the same along the two paths. In symbols, we can
write this result as
Wab,1 Wab,2, (8-2)
where the subscript ab indicates the initial and final points, respectively, and the
subscripts 1 and 2 indicate the path.
This result is powerful because it allows us to simplify difficult problems
when only a conservative force is involved. Suppose you need to calculate the
work done by a conservative force along a given path between two points, and
the calculation is difficult or even impossible without additional information.
You can find the work by substituting some other path between those two points
for which the calculation is easier and possible.
Proof of Equation 8-2
Figure 8-4bshows an arbitrary round trip for a particle that is acted upon by a single
force. The particle moves from an initial point ato point balong path 1 and then
back to point aalong path 2. The force does work on the particle as the particle
moves along each path. Without worrying about where positive work is done and
where negative work is done, let us just represent the work done from ato balong
path 1 as Wab,1 and the work done from bback to aalong path 2 as Wba,2. If the force
is conservative,then the net work done during the round trip must be zero:
Wab,1 Wba,2 0,
and thus
Wab,1 Wba,2. (8-3)
In words, the work done along the outward path must be the negative of the work
done along the path back.
Let us now consider the work Wab,2 done on the particle by the force when
the particle moves from ato balong path 2, as indicated in Fig. 8-4a. If the force is
conservative, that work is the negative of Wba,2:
Wab,2 Wba,2. (8-4)
Substituting Wab,2 for Wba,2 in Eq. 8-3, we obtain
Wab,1 Wab,2,
which is what we set out to prove.
Determining Potential Energy Values
Here we find equations that give the value of the two types of potential energy
discussed in this chapter: gravitational potential energy and elastic potential
energy. However, first we must find a general relation between a conservative
force and the associated potential energy.
Consider a particle-like object that is part of a system in which a conservative
force acts. When that force does work Won the object, the change Uin
the potential energy associated with the system is the negative of the work done.
We wrote this fact as Eq. 8-1 (UW). For the most general case, in which the
force may vary with position, we may write the work Was in Eq. 7-32:
(8-5)
This equation gives the work done by the force when the object moves from
point xito point xf, changing the configuration of the system. (Because the
force is conservative, the work is the same for all paths between those two
points.)
Wxf
xi
F(x)dx.
F
:
181
8-1 POTENTIAL ENERGY
Sample Problem 8.01 Equivalent paths for calculating work, slippery cheese
The main lesson of this sample problem is this: It is perfectly
all right to choose an easy path instead of a hard path.
Figure 8-5ashows a 2.0 kg block of slippery cheese that
slides along a frictionless track from point ato point b.The
cheese travels through a total distance of 2.0 m along the
track, and a net vertical distance of 0.80 m. How much work is
done on the cheese by the gravitational force during the slide?
KEY IDEAS
(1) We cannot calculate the work by using Eq. 7-12 (Wg
mgd cos f). The reason is that the angle fbetween the
directions of the gravitational force and the displacement
varies along the track in an unknown way. (Even if we did
know the shape of the track and could calculate falong it,
the calculation could be very difficult.) (2) Because is a
conservative force, we can find the work by choosing some
other path between aand bone that makes the calcula-
tion easy.
Calculations: Let us choose the dashed path in Fig. 8-5b;it
consists of two straight segments. Along the horizontal seg-
ment, the angle fis a constant 90. Even though we do not
know the displacement along that horizontal segment, Eq. 7-12
tells us that the work Whdone there is
Whmgd cos 900.
Along the vertical segment, the displacement dis 0.80 m
and, with and both downward, the angle fis a constantd
:
F
:
g
F
:
g
d
:F
:
g
Additional examples, video, and practice available at WileyPLUS
a
(
a
)(
b
)
b
a
b
The gravitational force is conservative.
Any choice of path between the points
gives the same amount of work.
Figure 8-5 (a) A block of cheese slides along a frictionless track
from point ato point b.(b) Finding the work done on the cheese by
the gravitational force is easier along the dashed path than along
the actual path taken by the cheese; the result is the same for
both paths.
vertical part of the dashed path,
Wvmgd cos 0
(2.0 kg)(9.8 m/s2)(0.80 m)(1) 15.7 J.
The total work done on the cheese by as the cheese
moves from point ato point balong the dashed path is then
WWhWv015.7 J 16 J. (Answer)
This is also the work done as the cheese slides along the
track from ato b.
F
:
g
0. Thus, Eq. 7-12 gives us, for the work Wvdone along the
Substituting Eq. 8-5 into Eq. 8-1, we find that the change in potential energy
due to the change in configuration is, in general notation,
(8-6)
Gravitational Potential Energy
We first consider a particle with mass mmoving vertically along a yaxis (the
positive direction is upward). As the particle moves from point yito point yf,
the gravitational force does work on it.To find the corresponding change in
the gravitational potential energy of the particleEarth system, we use Eq. 8-6
with two changes: (1) We integrate along the yaxis instead of the xaxis, because
the gravitational force acts vertically. (2) We substitute mg for the force symbol F,
because has the magnitude mg and is directed down the yaxis.We then have
which yields
Umg(yfyi)mg y. (8-7)
Only changes Uin gravitational potential energy (or any other type of
potential energy) are physically meaningful. However, to simplify a calculation or
a discussion, we sometimes would like to say that a certain gravitational potential
value Uis associated with a certain particleEarth system when the particle is at
a certain height y.To do so, we rewrite Eq. 8-7 as
UUimg(yyi). (8-8)
Then we take Uito be the gravitational potential energy of the system when it is
in a reference configuration in which the particle is at a reference point yi.
Usually we take Ui0 and yi0. Doing this changes Eq. 8-8 to
U(y)mgy (gravitational potential energy). (8-9)
This equation tells us:
U
yf
yi
(mg)dy mg yf
yi
dy mg
y
yf
yi
,
F
:
g
F
:
g
U
xf
xi
F(x)dx.
182 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Elastic Potential Energy
We next consider the blockspring system shown in Fig. 8-3, with the block
moving on the end of a spring of spring constant k. As the block moves from
point xito point xf, the spring force Fxkx does work on the block.To find the
corresponding change in the elastic potential energy of the blockspring system,
we substitute kx for F(x) in Eq. 8-6.We then have
or (8-10)
To associate a potential energy value Uwith the block at position x,we
choose the reference configuration to be when the spring is at its relaxed length
and the block is at xi0. Then the elastic potential energy Uiis 0, and Eq. 8-10
U1
2kxf
21
2kxi
2.
U
xf
xi
(kx)dx kxf
xi
xdx1
2k
x2
xf
xi
,
The gravitational potential energy associated with a particle Earth system
depends only on the vertical position y(or height) of the particle relative to the
reference position y0, not on the horizontal position.
183
8-1 POTENTIAL ENERGY
Checkpoint 2
A particle is to move along an xaxis from x0 to x1while a conser-
vative force, directed along the xaxis, acts on the particle.The figure
shows three situations in which the xcomponent of that force varies
with x.The force has the same maximum magnitude F1in all three sit-
uations.Rank the situations according to the change in the associated
potential energy during the particle’s motion,most positive first.
F1F1
F1
x1
x1
x1
(
1
)(
2
)(
3
)
the ground, (3) at the limb, and (4) 1.0 m above the limb?
Take the gravitational potential energy to be zero at y0.
KEY IDEA
Once we have chosen the reference point for y0, we can
calculate the gravitational potential energy Uof the system
relative to that reference point with Eq. 8-9.
Calculations: For choice (1) the sloth is at y5.0 m, and
Umgy (2.0 kg)(9.8 m/s2)(5.0 m)
98 J. (Answer)
For the other choices, the values of Uare
(2) Umgy mg(2.0 m) 39 J,
(3) Umgy mg(0) 0J,
(4) Umgy mg(1.0 m)
19.6 J 20 J. (Answer)
(b) The sloth drops to the ground. For each choice of refer-
ence point, what is the change Uin the potential energy of
the slothEarth system due to the fall?
KEY IDEA
The change in potential energy does not depend on the
choice of the reference point for y0; instead, it depends
on the change in height y.
Calculation: For all four situations, we have the same y
5.0 m.Thus,for (1) to (4), Eq. 8-7 tells us that
Umg y(2.0 kg)(9.8 m/s2)(5.0 m)
98 J. (Answer)
Sample Problem 8.02 Choosing reference level for gravitational potential energy, sloth
Here is an example with this lesson plan: Generally you can
choose any level to be the reference level, but once chosen,
be consistent. A 2.0 kg sloth hangs 5.0 m above the ground
(Fig. 8-6).
(a) What is the gravitational potential energy Uof the
slothEarth system if we take the reference point y0 to be
(1) at the ground, (2) at a balcony floor that is 3.0 m above
Additional examples, video, and practice available at WileyPLUS
0 –3 –5 –6
3 0 –2 –3
5 2 0
6 3 1 0
(1) (2) (3) (4)
Figure 8-6 Four choices of reference point y0. Each yaxis is marked
in units of meters.The choice affects the value of the potential energy
Uof the slothEarth system. However, it does not affect the change
Uin potential energy of the system if the sloth moves by, say, falling.
becomes
which gives us
(elastic potential energy). (8-11)U(x)1
2kx2
U01
2kx20,
Conservation of Mechanical Energy
The mechanical energy Emec of a system is the sum of its potential energy Uand
the kinetic energy Kof the objects within it:
Emec KU(mechanical energy). (8-12)
In this module, we examine what happens to this mechanical energy when only
conservative forces cause energy transfers within the systemthat is, when
frictional and drag forces do not act on the objects in the system. Also, we shall
assume that the system is isolated from its environment; that is, no external force
from an object outside the system causes energy changes inside the system.
When a conservative force does work Won an object within the system, that
force transfers energy between kinetic energy Kof the object and potential
energy Uof the system. From Eq. 7-10, the change Kin kinetic energy is
KW(8-13)
and from Eq. 8-1, the change Uin potential energy is
UW. (8-14)
Combining Eqs. 8-13 and 8-14, we find that
KU. (8-15)
In words, one of these energies increases exactly as much as the other decreases.
We can rewrite Eq. 8-15 as
K2K1(U2U1), (8-16)
where the subscripts refer to two different instants and thus to two different
arrangements of the objects in the system. Rearranging Eq. 8-16 yields
K2U2K1U1(conservation of mechanical energy). (8-17)
In words, this equation says:
the sum of K and U for
any state of a system
the sum of K and U for
any other state of the system
,
184 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
8-2 CONSERVATION OF MECHANICAL ENERGY
After reading this module, you should be able to . . .
8.05 After first clearly defining which objects form a system,
identify that the mechanical energy of the system is the
sum of the kinetic energies and potential energies of those
objects.
8.06 For an isolated system in which only conservative forces
act, apply the conservation of mechanical energy to relate
the initial potential and kinetic energies to the potential and
kinetic energies at a later instant.
Learning Objectives
Key Ideas
The mechanical energy Emec of a system is the sum of its
kinetic energy Kand potential energy U:
Emec KU.
An isolated system is one in which no external force causes
energy changes. If only conservative forces do work within
an isolated system, then the mechanical energy Emec of the
system cannot change. This principle of conservation of
mechanical energy is written as
K2U2K1U1,
in which the subscripts refer to different instants during an
energy transfer process. This conservation principle can also
be written as
Emec KU0.
©AP/Wide World Photos
In olden days, a person would be tossed
via a blanket to be able to see farther
over the flat terrain. Nowadays, it is
done just for fun. During the ascent of
the person in the photograph, energy is
transferred from kinetic energy to gravita-
tional potential energy. The maximum
height is reached when that transfer is
complete. Then the transfer is reversed
during the fall.
when the system is isolated and only conservative forces act on the objects in the
system. In other words:
185
8-2 CONSERVATION OF MECHANICAL ENERGY
In an isolated system where only conservative forces cause energy changes, the
kinetic energy and potential energy can change, but their sum, the mechanical
energy Emec of the system, cannot change.
When the mechanical energy of a system is conserved, we can relate the sum of kinetic
energy and potential energy at one instant to that at another instant without consider-
ing the intermediate motion and without finding the work done by the forces involved.
This result is called the principle of conservation of mechanical energy. (Now you
can see where conservative forces got their name.) With the aid of Eq. 8-15, we
can write this principle in one more form, as
Emec KU0. (8-18)
The principle of conservation of mechanical energy allows us to solve
problems that would be quite difficult to solve using only Newton’s laws:
Figure 8-7 shows an example in which the principle of conservation of
mechanical energy can be applied: As a pendulum swings, the energy of the
Figure 8-7 A pendulum, with its mass
concentrated in a bob at the lower end,
swings back and forth. One full cycle of
the motion is shown. During the cycle the
values of the potential and kinetic ener-
gies of the pendulumEarth system vary
as the bob rises and falls, but the mechani-
cal energy Emec of the system remains
constant. The energy Emec can be
described as continuously shifting between
the kinetic and potential forms. In stages
(a) and (e), all the energy is kinetic energy.
The bob then has its greatest speed and is
at its lowest point. In stages (c) and (g), all
the energy is potential energy. The bob
then has zero speed and is at its highest
point. In stages (b), (d), ( f), and (h), half
the energy is kinetic energy and half is
potential energy. If the swinging involved
a frictional force at the point where the
pendulum is attached to the ceiling, or a
drag force due to the air, then Emec would
not be conserved, and eventually the
pendulum would stop.
(a)
KU
(b)
KU
(c)
KU
(d)
KU
(e)
KU
(h)
KU
(f)
KU
(g)
KU
v = +vmax
v = 0
v = –vmax
v = 0
v
v
v
v
v
v
All potential
energy
All potential
energy
The total energy
does not change
(it is conserved).
All kinetic energy
All kinetic energy
pendulumEarth system is transferred back and forth between kinetic energy K
and gravitational potential energy U, with the sum KUbeing constant. If we
know the gravitational potential energy when the pendulum bob is at its highest
point (Fig. 8-7c), Eq. 8-17 gives us the kinetic energy of the bob at the lowest
point (Fig.8-7e).
For example, let us choose the lowest point as the reference point, with the
gravitational potential energy U20. Suppose then that the potential energy at
the highest point is U120 J relative to the reference point. Because the bob mo-
mentarily stops at its highest point, the kinetic energy there is K10. Putting these
values into Eq. 8-17 gives us the kinetic energy K2at the lowest point:
K20020 J or K220 J.
Note that we get this result without considering the motion between the highest
and lowest points (such as in Fig. 8-7d) and without finding the work done by any
forces involved in the motion.
186 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Checkpoint 3
The figure shows four
situationsone in
which an initially sta-
tionary block is dropped
and three in which the
block is allowed to slide
down frictionless ramps.
(a) Rank the situations
according to the kinetic energy of the block at point B,greatest first.(b) Rank them
according to the speed of the block at point B,greatest first.
A
B B B B
(
1
)(
2
)(
3
)(
4
)
System: Because the only force doing work on the child
is the gravitational force, we choose the childEarth system
as our system, which we can take to be isolated.
Thus, we have only a conservative force doing work in
an isolated system, so we can use the principle of conserva-
tion of mechanical energy.
Calculations: Let the mechanical energy be Emec,twhen the
child is at the top of the slide and Emec,bwhen she is at the
bottom.Then the conservation principle tells us
Emec,bEmec,t. (8-19)
Sample Problem 8.03 Conservation of mechanical energy, water slide
The huge advantage of using the conservation of energy in-
stead of Newton’s laws of motion is that we can jump from
the initial state to the final state without considering all the
intermediate motion. Here is an example. In Fig. 8-8, a child
of mass mis released from rest at the top of a water slide,
at height h8.5 m above the bottom of the slide.
Assuming that the slide is frictionless because of the water
on it, find the child’s speed at the bottom of the slide.
KEY IDEAS
(1) We cannot find her speed at the bottom by using her ac-
celeration along the slide as we might have in earlier chap-
ters because we do not know the slope (angle) of the slide.
However, because that speed is related to her kinetic en-
ergy, perhaps we can use the principle of conservation of
mechanical energy to get the speed. Then we would not
need to know the slope. (2) Mechanical energy is conserved
in a system if the system is isolated and if only conservative
forces cause energy transfers within it. Let’s check.
Forces: Two forces act on the child. The gravitational
force, a conservative force, does work on her. The normal
force on her from the slide does no work because its direc-
tion at any point during the descent is always perpendicular
to the direction in which the child moves.
Figure 8-8 A child slides down a water slide as she descends a
height h.
h
The total mechanical
energy at the top
is equal to the total
at the bottom.
Reading a Potential Energy Curve
Once again we consider a particle that is part of a system in which a conservative
force acts. This time suppose that the particle is constrained to move along an
xaxis while the conservative force does work on it.We want to plot the potential
energy U(x) that is associated with that force and the work that it does, and then
we want to consider how we can relate the plot back to the force and to the kinetic
energy of the particle. However, before we discuss such plots, we need one more
relationship between the force and the potential energy.
Finding the Force Analytically
Equation 8-6 tells us how to find the change Uin potential energy between two
points in a one-dimensional situation if we know the force F(x). Now we want to
187
8-3 READING A POTENTIAL ENERGY CURVE
To show both kinds of mechanical energy, we have
KbUbKtUt, (8-20)
or
Dividing by mand rearranging yield
Putting vt0 and ytybhleads to
(Answer)13 m/s.
vb12gh 1(2)(9.8 m/s2)(8.5 m)
vb
2vt
22g(ytyb).
1
2mvb
2mgyb1
2mvt
2mgyt.
Additional examples, video, and practice available at WileyPLUS
8-3 READING A POTENTIAL ENERGY CURVE
After reading this module, you should be able to . . .
8.07 Given a particle’s potential energy as a function of its
position x, determine the force on the particle.
8.08 Given a graph of potential energy versus x, determine
the force on a particle.
8.09 On a graph of potential energy versus x, superimpose a
line for a particle’s mechanical energy and determine the
particle’s kinetic energy for any given value of x.
8.10 If a particle moves along an xaxis, use a potential-
energy graph for that axis and the conservation of mechan-
ical energy to relate the energy values at one position to
those at another position.
8.11 On a potential-energy graph, identify any turning points
and any regions where the particle is not allowed because
of energy requirements.
8.12 Explain neutral equilibrium, stable equilibrium, and
unstable equilibrium.
Learning Objectives
Key Ideas
If we know the potential energy function U(x)for a system
in which a one-dimensional force F(x)acts on a particle, we
can find the force as
If U(x)is given on a graph, then at any value of x, the force
F(x)is the negative of the slope of the curve there and the
F(x) dU(x)
dx .
kinetic energy of the particle is given by
K(x)Emec U(x),
where Emec is the mechanical energy of the system.
A turning point is a point xat which the particle reverses its
motion (there, K0).
The particle is in equilibrium at points where the slope of
the U(x)curve is zero (there, F(x)0).
This is the same speed that the child would reach if she fell
8.5 m vertically. On an actual slide, some frictional forces
would act and the child would not be moving quite so fast.
Comments: Although this problem is hard to solve directly
with Newton’s laws, using conservation of mechanical en-
ergy makes the solution much easier. However, if we were
asked to find the time taken for the child to reach the bot-
tom of the slide, energy methods would be of no use; we
would need to know the shape of the slide, and we would
have a difficult problem.
go the other way; that is, we know the potential energy function U(x) and want
to find the force.
For one-dimensional motion, the work Wdone by a force that acts on a parti-
cle as the particle moves through a distance xis F(x)x. We can then write
Eq. 8-1 as
U(x)WF(x)x. (8-21)
Solving for F(x) and passing to the differential limit yield
(one-dimensional motion), (8-22)
which is the relation we sought.
We can check this result by putting , which is the elastic poten-
tial energy function for a spring force. Equation 8-22 then yields, as expected,
F(x)kx, which is Hooke’s law. Similarly, we can substitute U(x)mgx,
which is the gravitational potential energy function for a particleEarth system,
with a particle of mass mat height xabove Earth’s surface. Equation 8-22 then
yields Fmg, which is the gravitational force on the particle.
The Potential Energy Curve
Figure 8-9ais a plot of a potential energy function U(x) for a system in which a
particle is in one-dimensional motion while a conservative force F(x) does work
on it.We can easily find F(x) by (graphically) taking the slope of the U(x) curve at
various points. (Equation 8-22 tells us that F(x) is the negative of the slope of the
U(x) curve.) Figure 8-9bis a plot of F(x) found in this way.
Turning Points
In the absence of a nonconservative force, the mechanical energy Eof a system
has a constant value given by
U(x)K(x)Emec. (8-23)
Here K(x) is the kinetic energy function of a particle in the system (this K(x)
gives the kinetic energy as a function of the particle’s location x). We may
rewrite Eq. 8-23 as
K(x)Emec U(x). (8-24)
Suppose that Emec (which has a constant value, remember) happens to be 5.0 J. It
would be represented in Fig. 8-9cby a horizontal line that runs through the value
5.0 J on the energy axis. (It is, in fact, shown there.)
Equation 8-24 and Fig. 8-9dtell us how to determine the kinetic energy Kfor
any location xof the particle: On the U(x) curve, find Ufor that location xand
then subtract Ufrom Emec. In Fig. 8-9efor example, if the particle is at any point
to the right of x5, then K1.0 J.The value of Kis greatest (5.0 J) when the parti-
cle is at x2and least (0 J) when the particle is at x1.
Since Kcan never be negative (because v2is always positive), the particle can
never move to the left of x1, where Emec Uis negative. Instead, as the particle
moves toward x1from x2,Kdecreases (the particle slows) until K0 at x1(the
particle stops there).
Note that when the particle reaches x1, the force on the particle, given by
Eq. 8-22, is positive (because the slope dU/dx is negative). This means that the
particle does not remain at x1but instead begins to move to the right, opposite its
earlier motion. Hence x1is a turning point, a place where K0 (because UE)
and the particle changes direction. There is no turning point (where K0) on
the right side of the graph. When the particle heads to the right, it will continue
indefinitely.
U(x)1
2kx2
F(x)dU(x)
dx
188 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
189
8-3 READING A POTENTIAL ENERGY CURVE
6
5
4
3
2
1
U(J)
x
x2
x1x3x4x5
(a)
U(x)
+
x
(b)
x2
x1x3x4x5
F (N)
Mild force, –x direction
Strong force, +xdirection
This is a plot of the potential
energy U versus position x.
Force is equal to the negative of
the slope of the U(x) plot.
6
5
4
3
2
1
U (J), Emec (J)
x
x2
x1x3x4x5
(c)
Emec = 5.0 J
U(x)
The flat line shows a given value of
the total mechanical energy Emec.
The difference between the total energy
and the potential energy is the
kinetic energy K.
6
5
4
3
2
1
U (J), Emec (J)
x
x2
x1x3x4x5
(d)
Emec = 5.0 J
U(x)
K
6
5
4
3
2
1
x
(f)x2
x1x3x4x5
U (J), Emec (J)
At this position, K is greatest and
the particle is moving the fastest.
At this position, K is zero (a turning point).
The particle cannot go farther to the left.
For either of these three choices for Emec,
the particle is trapped (cannot escape
left or right).
6
5
4
3
2
1
U (J), Emec (J)
x
x2
x1x3x4x5
(e)
Emec = 5.0 J
K = 1.0 J at x > x5
K = 5.0 J at x2
A
Figure 8-9 (a) A plot of U(x), the potential energy function of a system containing a particle confined to move along an xaxis. There is no
friction, so mechanical energy is conserved. (b) A plot of the force F(x) acting on the particle, derived from the potential energy plot by
taking its slope at various points. (c)–(e) How to determine the kinetic energy. ( f) The U(x) plot of (a) with three possible values of Emec
shown. In WileyPLUS, this figure is available as an animation with voiceover.
Equilibrium Points
Figure 8-9fshows three different values for Emec superposed on the plot of the
potential energy function U(x) of Fig. 8-9a. Let us see how they change the situa-
tion. If Emec 4.0 J (purple line), the turning point shifts from x1to a point
between x1and x2. Also, at any point to the right of x5, the system’s mechanical
energy is equal to its potential energy; thus, the particle has no kinetic energy and
(by Eq. 8-22) no force acts on it, and so it must be stationary. A particle at such a
position is said to be in neutral equilibrium. (A marble placed on a horizontal
tabletop is in that state.)
If Emec 3.0 J (pink line), there are two turning points: One is between
x1and x2, and the other is between x4and x5. In addition, x3is a point at which
K0. If the particle is located exactly there, the force on it is also zero, and the
particle remains stationary. However, if it is displaced even slightly in either
direction, a nonzero force pushes it farther in the same direction, and the particle
continues to move. A particle at such a position is said to be in unstable equilib-
rium. (A marble balanced on top of a bowling ball is an example.)
Next consider the particle’s behavior if Emec 1.0 J (green line). If we place it
at x4,it is stuck there. It cannot move left or right on its own because to do so would
require a negative kinetic energy. If we push it slightly left or right, a restoring force
appears that moves it back to x4.A particle at such a position is said to be in stable
equilibrium. (A marble placed at the bottom of a hemispherical bowl is an example.)
If we place the particle in the cup-like potential well centered at x2,it is between two
turning points. It can still move somewhat, but only partway to x1or x3.
190 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Checkpoint 4
The figure gives the potential energy function
U(x) for a system in which a particle is in one-
dimensional motion. (a) Rank regions AB,BC, and
CD according to the magnitude of the force on the
particle, greatest first. (b) What is the direction of
the force when the particle is in region AB?
Calculations: At ,the particle has kinetic energy
Because the potential energy there is , the mechanical
energy is
.
This value for is plotted as a horizontal line in Fig. 8-10a.
From that figure we see that at , the potential
energy is . The kinetic energy is the difference
between and :
.
Because , we find
. (Answer)
(b) Where is the particle’s turning point located?
v13.0 m/s
K11
2mv1
2
K1Emec U116.0 J 7.0 J 9.0 J
U1
Emec
K1
U17.0 J
x4.5 m
Emec
Emec K0U016.0 J 016.0 J
U0
16.0 J.
K01
2mv2
01
2(2.00 kg)(4.00 m/s)2
x6.5 m
Sample Problem 8.04 Reading a potential energy graph
A 2.00 kg particle moves along an xaxis in one-dimensional
motion while a conservative force along that axis acts on it.
The potential energy U(x) associated with the force is plot-
ted in Fig. 8-10a. That is, if the particle were placed at any
position between and , it would have the
plotted value of U. At , the particle has velocity
.
(a) From Fig. 8-10a, determine the particle’s speed at
.
KEY IDEAS
(1) The particle’s kinetic energy is given by Eq. 7-1
( ). (2) Because only a conservative force acts on
the particle, the mechanical energy is con-
served as the particle moves. (3) Therefore, on a plot of U(x)
such as Fig. 8-10a, the kinetic energy is equal to the differ-
ence between and U.Emec
Emec (KU)
K1
2mv2
x14.5 m
i
ˆ
v0
:(4.00 m/s)
x6.5 m
x7.00 mx0
Additional examples, video, and practice available at WileyPLUS
191
8-4 WORK DONE ON A SYSTEM BY AN EXTERNAL FORCE
KEY IDEA
The turning point is where the force momentarily stops and
then reverses the particle’s motion. That is, it is where the
particle momentarily has and thus .
Calculations: Because Kis the difference between
, we want the point in Fig. 8-10awhere the plot of
Urises to meet the horizontal line of , as shown in Fig.
8-10b. Because the plot of Uis a straight line in Fig. 8-10b,
we can draw nested right triangles as shown and then write
the proportionality of distances
,
which gives us .Thus, the turning point is at
. (Answer)
(c) Evaluate the force acting on the particle when it is in the
region .
KEY IDEA
The force is given by Eq. 8-22 (F(x)dU(x)/dx):The force
is equal to the negative of the slope on a graph of U(x).
Calculations: For the graph of Fig. 8-10b, we see that for
the range the force is
. (Answer)F 20 J 7.0 J
1.0 m 4.0 m 4.3 N
1.0 m x4.0 m
1.9 m x4.0 m
x4.0 m d1.9 m
d2.08 m
16 7.0
d20 7.0
4.0 1.0
Emec
Emec and U
K0v0
Thus, the force has magnitude 4.3 N and is in the positive
direction of the xaxis. This result is consistent with the fact
that the initially leftward-moving particle is stopped by the
force and then sent rightward.
Figure 8-10 (a) A plot of potential energy Uversus position x.
(b) A section of the plot used to find where the particle turns
around.
Kinetic energy is the difference
between the total energy and
the potential energy.
K1
K0
Emec = 16 J
20
16
7
01 4567
x (m)
(a)
U ( J)
Turning point
20
16
714
x (m)
d
(b)
U ( J)
The kinetic energy is zero
at the turning point (the
particle speed is zero).
8-4 WORK DONE ON A SYSTEM BY AN EXTERNAL FORCE
After reading this module, you should be able to . . .
8.13 When work is done on a system by an external force
with no friction involved, determine the changes in kinetic
energy and potential energy.
8.14 When work is done on a system by an external force
with friction involved, relate that work to the changes in
kinetic energy, potential energy, and thermal energy.
Work Wis energy transferred to or from a system by means
of an external force acting on the system.
When more than one force acts on a system, their net work
is the transferred energy.
When friction is not involved, the work done on the system
and the change Emec in the mechanical energy of the system
are equal:
WEmec KU.
When a kinetic frictional force acts within the system, then
the thermal energy Eth of the system changes. (This energy is
associated with the random motion of atoms and molecules
in the system.) The work done on the system is then
WEmec Eth.
The change Eth is related to the magnitude fkof the frictional
force and the magnitude dof the displacement caused by the
external force by
Eth fkd.
Learning Objectives
Key Ideas
Work Done on a System by an External Force
In Chapter 7, we defined work as being energy transferred to or from an object
by means of a force acting on the object.We can now extend that definition to an
external force acting on a system of objects.
192 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Work is energy transferred to or from a system by means of an external force
acting on that system.
Figure 8-11arepresents positive work (a transfer of energy to a system), and
Fig. 8-11brepresents negative work (a transfer of energy from a system). When
more than one force acts on a system, their net work is the energy transferred to
or from the system.
These transfers are like transfers of money to and from a bank account. If a
system consists of a single particle or particle-like object, as in Chapter 7, the
work done on the system by a force can change only the kinetic energy of the
system.The energy statement for such transfers is the workkinetic energy theo-
rem of Eq. 7-10 (KW); that is, a single particle has only one energy account,
called kinetic energy. External forces can transfer energy into or out of that
account. If a system is more complicated, however, an external force can change
other forms of energy (such as potential energy); that is, a more complicated
system can have multiple energy accounts.
Let us find energy statements for such systems by examining two basic situa-
tions, one that does not involve friction and one that does.
No Friction Involved
To compete in a bowling-ball-hurling contest, you first squat and cup your hands
under the ball on the floor.Then you rapidly straighten up while also pulling your
hands up sharply, launching the ball upward at about face level. During your
upward motion, your applied force on the ball obviously does work; that is, it is an
external force that transfers energy, but to what system?
To answer, we check to see which energies change. There is a change Kin
the ball’s kinetic energy and, because the ball and Earth become more sepa-
rated, there is a change Uin the gravitational potential energy of the
ballEarth system.To include both changes, we need to consider the ballEarth
system. Then your force is an external force doing work on that system, and the
work is
WKU, (8-25)
or WEmec (work done on system, no friction involved), (8-26)
where Emec is the change in the mechanical energy of the system. These two
equations, which are represented in Fig. 8-12, are equivalent energy statements
for work done on a system by an external force when friction is not involved.
Friction Involved
We next consider the example in Fig. 8-13a. A constant horizontal force pulls a
block along an xaxis and through a displacement of magnitude d, increasing the
block’s velocity from to . During the motion, a constant kinetic frictional
force from the floor acts on the block. Let us first choose the block as our
system and apply Newton’s second law to it. We can write that law for compo-
nents along the xaxis (Fnet,xmax) as
Ffkma. (8-27)
f
:
k
v
:
v
:
0
F
:
Positive W
S
ystem
(a)
Negative W
System
(
b
)
Figure 8-11 (a) Positive work Wdone on an
arbitrary system means a transfer of
energy to the system. (b) Negative work
Wmeans a transfer of energy from the
system.
WΔEmec =ΔK+ΔU
Ball–Earth
system
Your lifting force
transfers energy to
kinetic energy and
potential energy.
Figure 8-12 Positive work Wis done on a
system of a bowling ball and Earth, caus-
ing a change Emec in the mechanical
energy of the system, a change Kin the
ball’s kinetic energy, and a change Uin
the system’s gravitational potential energy.
Because the forces are constant, the acceleration is also constant. Thus, we can
use Eq. 2-16 to write
Solving this equation for a, substituting the result into Eq. 8-27, and rearranging
then give us
(8-28)
or,because for the block,
Fd Kfkd. (8-29)
In a more general situation (say, one in which the block is moving up a ramp), there
can be a change in potential energy. To include such a possible change, we general-
ize Eq. 8-29 by writing
Fd Emec fkd. (8-30)
By experiment we find that the block and the portion of the floor along
which it slides become warmer as the block slides. As we shall discuss in
Chapter 18, the temperature of an object is related to the object’s thermal energy
Eth (the energy associated with the random motion of the atoms and molecules in
the object). Here, the thermal energy of the block and floor increases because
(1) there is friction between them and (2) there is sliding. Recall that friction is
due to the cold-welding between two surfaces. As the block slides over the floor,
the sliding causes repeated tearing and re-forming of the welds between the
block and the floor, which makes the block and floor warmer. Thus, the sliding
increases their thermal energy Eth.
Through experiment, we find that the increase Eth in thermal energy is
equal to the product of the magnitudes fkand d:
Eth fkd(increase in thermal energy by sliding). (8-31)
Thus, we can rewrite Eq. 8-30 as
Fd Emec Eth. (8-32)
Fd is the work Wdone by the external force (the energy transferred by the
force), but on which system is the work done (where are the energy transfers made)?
To answer, we check to see which energies change. The block’s mechanical energy
F
:
1
2mv21
2mv0
2K
Fd 1
2mv21
2mv0
2fkd
v2v0
22ad.
a
:
193
8-4 WORK DONE ON A SYSTEM BY AN EXTERNAL FORCE
fk
v0v
F
d
x
(a)
The applied force supplies energy.
The frictional force transfers some
of it to thermal energy.
(b)
Block–floor
system
ΔEmec
ΔEth
W
So, the work done by the applied
force goes into kinetic energy
and also thermal energy.
Figure 8-13 (a) A block is pulled across a floor by force while a kinetic frictional
force opposes the motion. The block has velocity at the start of a displacement
and velocity at the end of the displacement. (b) Positive work Wis done on the
blockfloor system by force , resulting in a change Emec in the block’s mechanical
energy and a change Eth in the thermal energy of the block and floor.
F
:
v
:
d
:
v
:
0
f
:
k
F
:
194 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Checkpoint 5
In three trials,a block is pushed
by a horizontal applied force
across a floor that is not friction-
less,as in Fig. 8-13a.The magni-
tudes Fof the applied force and
the results of the pushing on the
block’s speed are given in the
table. In all three trials, the block is pushed through the same distance d. Rank the
three trials according to the change in the thermal energy of the block and floor that
occurs in that distance d, greatest first.
Trial FResult on Block’s Speed
a 5.0 N decreases
b 7.0 N remains constant
c 8.0 N increases
be friction and a change Eth in thermal energy of the crate
and the floor. Therefore, the system on which the work is
done is the cratefloor system, because both energy
changes occur in that system.
(b) What is the increase Eth in the thermal energy of the
crate and floor?
KEY IDEA
We can relate Eth to the work Wdone by with the energy
statement of Eq. 8-33 for a system that involves friction:
WEmec Eth. (8-34)
Calculations: We know the value of Wfrom (a). The
change Emec in the crate’s mechanical energy is just the
change in its kinetic energy because no potential energy
changes occur,so we have
Substituting this into Eq.8-34 and solving for Eth,we find
(Answer)
Without further experiments, we cannot say how much of
this thermal energy ends up in the crate and how much in
the floor.We simply know the total amount.
22.2 J 22 J.
20 J 1
2(14 kg)[(0.20 m/s)2(0.60 m/s)2]
Eth W(1
2mv21
2mv0
2)W1
2m(v2v0
2)
Emec K1
2mv21
2mv0
2.
F
:
Sample Problem 8.05 Work, friction, change in thermal energy, cabbage heads
A food shipper pushes a wood crate of cabbage heads (total
mass m14 kg) across a concrete floor with a constant
horizontal force of magnitude 40 N. In a straight-line dis-
placement of magnitude d0.50 m, the speed of the crate
decreases from v00.60 m/s to v0.20 m/s.
(a) How much work is done by force , and on what system
does it do the work?
KEY IDEA
Because the applied force is constant, we can calculate
the work it does by using Eq. 7-7 ( ).
Calculation: Substituting given data, including the fact that
force and displacement are in the same direction, we
find
WFd cos f(40 N)(0.50 m) cos 0
20 J. (Answer)
Reasoning: To determine the system on which the work is
done, let’s check which energies change. Because the crate’s
speed changes, there is certainly a change Kin the crate’s
kinetic energy. Is there friction between the floor and the
crate, and thus a change in thermal energy? Note that and
the crate’s velocity have the same direction.Thus, if there is
no friction, then should be accelerating the crate to a
greater speed. However, the crate is slowing, so there must
F
:
F
:
d
:
F
:
WFd cos
F
:
F
:
F
:
Additional examples, video, and practice available at WileyPLUS
changes, and the thermal energies of the block and floor also change. Therefore, the
work done by force is done on the blockfloor system.That work is
WEmec Eth (work done on system, friction involved). (8-33)
This equation, which is represented in Fig. 8-13b, is the energy statement for the
work done on a system by an external force when friction is involved.
F
:
Conservation of Energy
We now have discussed several situations in which energy is transferred to or
from objects and systems, much like money is transferred between accounts.
In each situation we assume that the energy that was involved could always be
accounted for; that is, energy could not magically appear or disappear. In more
formal language, we assumed (correctly) that energy obeys a law called the law of
conservation of energy, which is concerned with the total energy Eof a system.
That total is the sum of the system’s mechanical energy, thermal energy, and any
type of internal energy in addition to thermal energy. (We have not yet discussed
other types of internal energy.) The law states that
195
8-5 CONSERVATION OF ENERGY
The total energy Eof a system can change only by amounts of energy that are
transferred to or from the system.
8-5 CONSERVATION OF ENERGY
After reading this module, you should be able to . . .
8.15 For an isolated system (no net external force), apply the
conservation of energy to relate the initial total energy
(energies of all kinds) to the total energy at a later instant.
8.16 For a nonisolated system, relate the work done on the
system by a net external force to the changes in the vari-
ous types of energies within the system.
8.17 Apply the relationship between average power, the
associated energy transfer, and the time interval in which
that transfer is made.
8.18 Given an energy transfer as a function of time (either as
an equation or a graph), determine the instantaneous
power (the transfer at any given instant).
Learning Objectives
The total energy Eof a system (the sum of its mechanical
energy and its internal energies, including thermal energy)
can change only by amounts of energy that are transferred to
or from the system. This experimental fact is known as the law
of conservation of energy.
If work Wis done on the system, then
WEEmec Eth Eint.
If the system is isolated (W0), this gives
Emec Eth Eint 0
and Emec,2 Emec,1 Eth Eint,
where the subscripts 1 and 2 refer to two different instants.
The power due to a force is the rate at which that force
transfers energy. If an amount of energy Eis transferred by
a force in an amount of time t, the average power of the
force is
The instantaneous power due to a force is
On a graph of energy Eversus time t, the power is the slope
of the plot at any given time.
PdE
dt .
Pavg E
t.
Key Ideas
The only type of energy transfer that we have considered is work Wdone on a
system by an external force.Thus, for us at this point, this law states that
WEEmec Eth Eint, (8-35)
where Emec is any change in the mechanical energy of the system, Eth is any
change in the thermal energy of the system, and Eint is any change in any
other type of internal energy of the system. Included in Emec are changes Kin
kinetic energy and changes Uin potential energy (elastic, gravitational, or any
other type we might find).
This law of conservation of energy is not something we have derived from
basic physics principles. Rather, it is a law based on countless experiments.
Scientists and engineers have never found an exception to it. Energy simply can-
not magically appear or disappear.
Isolated System
If a system is isolated from its environment, there can be no energy transfers to or
from it. For that case,the law of conservation of energy states:
196 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
The total energy Eof an isolated system cannot change.
Many energy transfers may be going on within an isolated systembetween,
say, kinetic energy and a potential energy or between kinetic energy and ther-
mal energy. However, the total of all the types of energy in the system cannot
change. Here again, energy cannot magically appear or disappear.
We can use the rock climber in Fig. 8-14 as an example, approximating
him, his gear, and Earth as an isolated system. As he rappels down the rock
face, changing the configuration of the system, he needs to control the transfer
of energy from the gravitational potential energy of the system. (That energy
cannot just disappear.) Some of it is transferred to his kinetic energy.
However, he obviously does not want very much transferred to that type or he
will be moving too quickly, so he has wrapped the rope around metal rings to
produce friction between the rope and the rings as he moves down.The sliding
of the rings on the rope then transfers the gravitational potential energy of the
system to thermal energy of the rings and rope in a way that he can control.
The total energy of the climbergearEarth system (the total of its gravita-
tional potential energy, kinetic energy, and thermal energy) does not change
during his descent.
For an isolated system, the law of conservation of energy can be written in
two ways. First, by setting W0 in Eq. 8-35, we get
Emec Eth Eint 0(isolated system). (8-36)
We can also let Emec Emec,2 Emec,1, where the subscripts 1 and 2 refer to two
different instantssay, before and after a certain process has occurred.Then Eq.
8-36 becomes
Emec,2 Emec,1 Eth Eint. (8-37)
Equation 8-37 tells us:
Figure 8-14 To descend, the rock climber
must transfer energy from the gravitational
potential energy of a system consisting of
him, his gear, and Earth. He has wrapped
the rope around metal rings so that the
rope rubs against the rings. This allows
most of the transferred energy to go to the
thermal energy of the rope and rings
rather than to his kinetic energy.
Tyler Stableford/The Image Bank/Getty Images
In an isolated system, we can relate the total energy at one instant to the total
energy at another instant without considering the energies at intermediate times.
This fact can be a very powerful tool in solving problems about isolated systems
when you need to relate energies of a system before and after a certain process
occurs in the system.
In Module 8-2, we discussed a special situation for isolated systemsnamely,
the situation in which nonconservative forces (such as a kinetic frictional force)
do not act within them. In that special situation, Eth and Eint are both zero, and
so Eq. 8-37 reduces to Eq. 8-18. In other words, the mechanical energy of an
isolated system is conserved when nonconservative forces do not act in it.
External Forces and Internal Energy Transfers
An external force can change the kinetic energy or potential energy of an object
without doing work on the objectthat is, without transferring energy to the
object. Instead, the force is responsible for transfers of energy from one type to
another inside the object.
Figure 8-15 shows an example.An initially stationary ice-skater pushes away
from a railing and then slides over the ice (Figs. 8-15aand b). Her kinetic energy
increases because of an external force on her from the rail. However, that force
does not transfer energy from the rail to her. Thus, the force does no work on
her. Rather, her kinetic energy increases as a result of internal transfers from the
biochemical energy in her muscles.
Figure 8-16 shows another example. An engine increases the speed of a car
with four-wheel drive (all four wheels are made to turn by the engine). During
the acceleration, the engine causes the tires to push backward on the road sur-
face. This push produces frictional forces that act on each tire in the forward
direction. The net external force from the road, which is the sum of these fric-
tional forces, accelerates the car, increasing its kinetic energy. However, does
not transfer energy from the road to the car and so does no work on the car.
Rather, the car’s kinetic energy increases as a result of internal transfers from the
energy stored in the fuel.
F
:
F
:f
:
F
:
197
8-5 CONSERVATION OF ENERGY
Figure 8-15 (a) As a skater pushes herself away from a railing, the force on her from
the railing is . (b) After the skater leaves the railing, she has velocity . (c) External
force acts on the skater, at angle fwith a horizontal xaxis. When the skater goes
through displacement , her velocity is changed from (0) to by the horizontal
component of .F
:v
:
v
:
0
d
:
F
:v
:
F
:
Ice
(a)
φ
F
φ
(c)
v0
x
F
v
d
(b)
v
Her push on the rail causes
a transfer of internal energy
to kinetic energy.
Figure 8-16 A vehicle accelerates to the
right using four-wheel drive. The road
exerts four frictional forces (two of them
shown) on the bottom surfaces of the tires.
Taken together, these four forces make up
the net external force acting on the car.F
:
acom
f f
In situations like these two, we can sometimes relate the external force on
an object to the change in the object’s mechanical energy if we can simplify the
situation. Consider the ice-skater example.During her push through distance din
Fig. 8-15c, we can simplify by assuming that the acceleration is constant, her
speed changing from v00 to v. (That is, we assume has constant magnitude F
and angle f.) After the push, we can simplify the skater as being a particle and
neglect the fact that the exertions of her muscles have increased the thermal
energy in her muscles and changed other physiological features. Then we can
apply Eq. 7-5 to write
KK0(Fcos f)d,
or KFd cos f. (8-38)
If the situation also involves a change in the elevation of an object, we can
include the change Uin gravitational potential energy by writing
UKFd cos f. (8-39)
The force on the right side of this equation does no work on the object but is still
responsible for the changes in energy shown on the left side.
Power
Now that you have seen how energy can be transferred from one type to another,
we can expand the definition of power given in Module 7-6. There power is
(1
2mv21
2mv0
2Fxd)
F
:
F
:
defined as the rate at which work is done by a force. In a more general sense,
power Pis the rate at which energy is transferred by a force from one type to
another. If an amount of energy Eis transferred in an amount of time t, the
average power due to the force is
(8-40)
Similarly, the instantaneous power due to the force is
(8-41)PdE
dt .
P
avg E
t.
198 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
on the glider to get it moving, a spring force does work on
it, transferring energy from the elastic potential energy of
the compressed spring to kinetic energy of the glider. The
spring force also pushes against a rigid wall. Because there
is friction between the glider and the ground-level track,
the sliding of the glider along that track section increases
their thermal energies.
System: Let’s take the system to contain all the interact-
ing bodies: glider, track, spring, Earth, and wall. Then, be-
cause all the force interactions are within the system, the
system is isolated and thus its total energy cannot change.
So, the equation we should use is not that of some external
force doing work on the system. Rather, it is a conservation
of energy.We write this in the form of Eq. 8-37:
Emec,2 Emec,1 Eth. (8-42)
This is like a money equation: The final money is equal to
the initial money minus the amount stolen away by a thief.
Here, the final mechanical energy is equal to the initial me-
chanical energy minus the amount stolen away by friction.
None has magically appeared or disappeared.
Calculations: Now that we have an equation, let’s find
distance L. Let subscript 1 correspond to the initial state
of the glider (when it is still on the compressed spring)
and subscript 2 correspond to the final state of the glider
(when it has come to rest on the ground-level track). For
both states, the mechanical energy of the system is the
sum of any potential energy and any kinetic energy.
We have two types of potential energy: the elastic po-
tential energy (Uekx2) associated with the compressed
1
2
Sample Problem 8.06 Lots of energies at an amusement park water slide
Figure 8-17 shows a water-slide ride in which a glider is shot
by a spring along a water-drenched (frictionless) track that
takes the glider from a horizontal section down to ground
level.As the glider then moves along ground-level track, it is
gradually brought to rest by friction. The total mass of the
glider and its rider is m200 kg, the initial compression of
the spring is d5.00 m, the spring constant is k3.20
103N/m, the initial height is h35.0 m, and the coefficient
of kinetic friction along the ground-level track is mk0.800.
Through what distance Ldoes the glider slide along the
ground-level track until it stops?
KEY IDEAS
Before we touch a calculator and start plugging numbers
into equations, we need to examine all the forces and then
determine what our system should be. Only then can we
decide what equation to write. Do we have an isolated sys-
tem (our equation would be for the conservation of en-
ergy) or a system on which an external force does work
(our equation would relate that work to the system’s
change in energy)?
Forces: The normal force on the glider from the track
does no work on the glider because the direction of this
force is always perpendicular to the direction of the
glider’s displacement. The gravitational force does work
on the glider, and because the force is conservative we can
associate a potential energy with it. As the spring pushes
spring and the gravitational potential energy (Ugmgy) as-
L
mk
m0
k
h
Figure 8-17 A spring-loaded amusement park water slide.
sociated with the glider’s elevation. For the latter, let’s take
ground level as the reference level. That means that the
glider is initially at height yhand finally at height y0.
In the initial state, with the glider stationary and ele-
vated and the spring compressed, the energy is
Emec,1 K1Ue1Ug1
0kd2mgh. (8-43)
1
2
Additional examples, video, and practice available at WileyPLUS
199
REVIEW & SUMMARY
Substituting Eqs. 8-43 through 8-45 into Eq. 8-42, we find
0kd2mgh mkmgL, (8-46)
and
L
69.3 m. (Answer)
Finally, note how algebraically simple our solution is. By
carefully defining a system and realizing that we have an
isolated system, we get to use the law of the conservation of
energy. That means we can relate the initial and final states
of the system with no consideration of the intermediate
states. In particular,we did not need to consider the glider as
it slides over the uneven track. If we had, instead, applied
Newton’s second law to the motion, we would have had to
know the details of the track and would have faced a far
more difficult calculation.
(3.20 103 N/m)(5.00 m)2
2(0.800)(200 kg)(9.8 m/s2)35 m
0.800
kd2
2mkmg h
mk
1
2
Conservative Forces A force is a conservative force if the net
work it does on a particle moving around any closed path, from an
initial point and then back to that point, is zero. Equivalently, a
force is conservative if the net work it does on a particle moving
between two points does not depend on the path taken by the par-
ticle. The gravitational force and the spring force are conservative
forces; the kinetic frictional force is a nonconservative force.
Potential Energy Apotential energy is energy that is associated
with the configuration of a system in which a conservative force acts.
When the conservative force does work Won a particle within the sys-
tem,the change Uin the potential energy of the system is
UW. (8-1)
If the particle moves from point xito point xf, the change in the
potential energy of the system is
(8-6)
Gravitational Potential Energy The potential energy asso-
ciated with a system consisting of Earth and a nearby particle is
gravitational potential energy. If the particle moves from height yi
to height yf, the change in the gravitational potential energy of the
particle–Earth system is
Umg(yfyi)mg y. (8-7)
If the reference point of the particle is set as yi0 and the cor-
responding gravitational potential energy of the system is set as
Ui0, then the gravitational potential energy Uwhen the parti-
U
xf
xi
F(x)dx.
Review & Summary
cle is at any height yis
U(y)mgy. (8-9)
Elastic Potential Energy Elastic potential energy is the
energy associated with the state of compression or extension of an
elastic object. For a spring that exerts a spring force Fkx when
its free end has displacement x, the elastic potential energy is
(8-11)
The reference configuration has the spring at its relaxed length, at
which x0 and U0.
Mechanical Energy The mechanical energy Emec of a system
is the sum of its kinetic energy Kand potential energy U:
Emec KU. (8-12)
An isolated system is one in which no external force causes energy
changes. If only conservative forces do work within an isolated sys-
tem, then the mechanical energy Emec of the system cannot change.
This principle of conservation of mechanical energy is written as
K2U2K1U1, (8-17)
in which the subscripts refer to different instants during an
energy transfer process. This conservation principle can also be
written as
Emec KU0. (8-18)
Potential Energy Curves If we know the potential energy
function U(x) for a system in which a one-dimensional force F(x)
U(x)1
2kx2.
In the final state, with the spring now in its relaxed state and
the glider again stationary but no longer elevated, the final
mechanical energy of the system is
Emec,2 K2Ue2Ug2
000. (8-44)
Let’s next go after the change Eth of the thermal energy of
the glider and ground-level track. From Eq. 8-31, we can
substitute for Eth with fkL(the product of the frictional
force magnitude and the distance of rubbing). From Eq. 6-2,
we know that fk mkFN, where FNis the normal force.
Because the glider moves horizontally through the region
with friction, the magnitude of FNis equal to mg (the up-
ward force matches the downward force). So, the friction’s
theft from the mechanical energy amounts to
Eth mkmgL. (8-45)
(By the way, without further experiments, we cannot say
how much of this thermal energy ends up in the glider and
how much in the track. We simply know the total amount.)
200 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
acts on a particle, we can find the force as
(8-22)
If U(x) is given on a graph, then at any value of x, the force F(x) is
the negative of the slope of the curve there and the kinetic energy
of the particle is given by
K(x)Emec U(x), (8-24)
where Emec is the mechanical energy of the system.A turning point
is a point xat which the particle reverses its motion (there, K0).
The particle is in equilibrium at points where the slope of the U(x)
curve is zero (there, F(x)0).
Work Done on a System by an External Force Work W
is energy transferred to or from a system by means of an external
force acting on the system. When more than one force acts on a
system, their net work is the transferred energy. When friction is
not involved, the work done on the system and the change Emec in
the mechanical energy of the system are equal:
WEmec KU. (8-26, 8-25)
When a kinetic frictional force acts within the system, then the ther-
mal energy Eth of the system changes. (This energy is associated with
the random motion of atoms and molecules in the system.) The
work done on the system is then
WEmec Eth. (8-33)
F(x) dU(x)
dx .
The change Eth is related to the magnitude fkof the frictional force
and the magnitude dof the displacement caused by the external
force by
Eth fkd. (8-31)
Conservation of Energy The total energy Eof a system
(the sum of its mechanical energy and its internal energies,
including thermal energy) can change only by amounts of energy
that are transferred to or from the system. This experimental fact
is known as the law of conservation of energy. If work Wis done
on the system, then
WEEmec Eth Eint. (8-35)
If the system is isolated (W0), this gives
Emec Eth Eint 0 (8-36)
and Emec,2 Emec,1 Eth Eint, (8-37)
where the subscripts 1 and 2 refer to two different instants.
Power The power due to a force is the rate at which that force
transfers energy. If an amount of energy Eis transferred by
a force in an amount of time t, the average power of the force is
(8-40)
The instantaneous power due to a force is
(8-41)PdE
dt .
Pavg E
t.
1In Fig. 8-18, a horizontally moving block can take three fric-
tionless routes, differing only in elevation, to reach the dashed
finish line. Rank the routes according to (a) the speed of the block
at the finish line and (b) the travel time of the block to the finish
line, greatest first.
tude of the force on the particle, greatest first. What value must
the mechanical energy Emec of the particle not exceed if the par-
ticle is to be (b) trapped in the potential well at the left, (c)
trapped in the potential well at the right, and (d) able to move
between the two potential wells but not to the right of point H?
For the situation of (d), in which of regions BC, DE, and FG will
the particle have (e) the greatest kinetic energy and (f) the least
speed?
3Figure 8-20 shows one direct
path and four indirect paths from
point ito point f. Along the direct
path and three of the indirect paths,
only a conservative force Fcacts on
a certain object. Along the fourth
indirect path, both Fcand a noncon-
servative force Fnc act on the object.
The change Emec in the object’s
mechanical energy (in joules) in going from ito fis indicated along
each straight-line segment of the indirect paths. What is Emec (a)
from ito falong the direct path and (b) due to Fnc along the one
path where it acts?
4In Fig. 8-21, a small, initially stationary block is released on a
frictionless ramp at a height of 3.0 m. Hill heights along the ramp
are as shown in the figure.The hills have identical circular tops, and
the block does not fly off any hill. (a) Which hill is the first the
block cannot cross? (b) What does the block do after failing
to cross that hill? Of the hills that the block can cross, on which hill-
Questions
(1)
Finish line
(2)
(3)
v
Figure 8-18 Question 1.
A B
C
D E F
G
H x
1
0
3
5
U(x) (J)
6
8
Figure 8-19 Question 2.
–30
40
32 10
2
–10
–6 –4
20
if
15 7
–30
Figure 8-20 Question 3.
2Figure 8-19 gives the potential energy function of a particle.
(a) Rank regions AB, BC, CD, and DE according to the magni-
descends, it pulls on a block via a
second rope, and the block slides
over a lab table. Again consider the
cylinderrodEarth system, similar
to that shown in Fig. 8-23b. Your
work on the system is 200 J. The sys-
tem does work of 60 J on the block.
Within the system, the kinetic
energy increases by 130 J and
the gravitational potential energy
decreases by 20 J. (a) Draw an “en-
ergy statement” for the system, as in
Fig. 8-23c. (b) What is the change in
the thermal energy within the system?
8In Fig. 8-25, a block slides along a track that descends through
distance h. The track is frictionless except for the lower section.
There the block slides to a stop in a certain distance Dbecause of
friction. (a) If we decrease h, will the block now slide to a stop in a
distance that is greater than, less than, or equal to D? (b) If, instead,
we increase the mass of the block, will the stopping distance now be
greater than, less than, or equal to D?
201
QUESTIONS
top is (c) the centripetal acceleration of the block greatest and (d) the
normal force on the block least?
(
1
) (
3
)(
2
)
Figure 8-26 Question 9.
hD
Figure 8-25 Question 8.
Figure 8-21 Question 4.
System's energies:
ΔK = +50 J
ΔUg = +20 J
ΔEth = ?
W = +100 J
(b)(a) (c)
Cylinder
Earth
Rope
Rod
System
Work W
A
B
C D
Figure 8-22 Question 5.
Figure 8-23 Question 6.
(
1
)
0.5 m
1.5 m
3.0 m
2.5 m
3.5 m
(2)
(3)
(4)
Cylinder
Rod
Block
Rope
Figure 8-24 Question 7.
5In Fig. 8-22, a block slides from Ato Calong a frictionless ramp,
and then it passes through horizontal region CD, where a frictional
force acts on it. Is the block’s kinetic energy increasing, decreasing,
or constant in (a) region AB, (b) region BC, and (c) region CD?
(d) Is the block’s mechanical energy increasing, decreasing, or
constant in those regions?
6In Fig. 8-23a, you pull upward on a rope that is attached to a
cylinder on a vertical rod. Because the cylinder fits tightly on the
rod, the cylinder slides along the rod with considerable friction.
Your force does work W100 J on the cylinderrodEarth
system (Fig. 8-23b).An “energy statement” for the system is shown
in Fig. 8-23c: the kinetic energy Kincreases by 50 J, and the gravita-
tional potential energy Ugincreases by 20 J.The only other change
in energy within the system is for the thermal energy Eth. What is
the change Eth?
10 Figure 8-27 shows three plums
that are launched from the same level
with the same speed. One moves
straight upward, one is launched at a
small angle to the vertical, and one is
launched along a frictionless incline.
Rank the plums according to their
speed when they reach the level of
the dashed line, greatest first.
11 When a particle moves from f
to iand from jto ialong the paths
shown in Fig. 8-28, and in the indi-
cated directions, a conservative
force does the indicated amounts
of work on it. How much work is
done on the particle by when the
particle moves directly from fto j?
F
:
F
:
(1) (2) (3)
Figure 8-27 Question 10.
f
i
j
20 J
20 J
Figure 8-28 Question 11.
7The arrangement shown in Fig. 8-24 is similar to that in
Question 6. Here you pull downward on the rope that is attached
to the cylinder, which fits tightly on the rod. Also, as the cylinder
9Figure 8-26 shows three situations involving a plane that is not
frictionless and a block sliding along the plane.The block begins with
the same speed in all three situations and slides until the kinetic fric-
tional force has stopped it. Rank the situations according to the in-
crease in thermal energy due to the sliding,greatest first.
•3 You drop a 2.00 kg book to a friend
who stands on the ground at distance
D10.0 m below. If your friend’s out-
stretched hands are at distance d1.50 m
above the ground (Fig. 8-30), (a) how
much work Wgdoes the gravitational
force do on the book as it drops to her
hands? (b) What is the change Uin the
gravitational potential energy of the
bookEarth system during the drop? If
the gravitational potential energy Uof
that system is taken to be zero at ground
level, what is U(c) when the book is re-
leased and (d) when it reaches her
hands? Now take Uto be 100 J at
ground level and again find (e) Wg,
(f) U, (g) Uat the release point, and
(h) Uat her hands.
•4 Figure 8-31 shows a ball with mass
m0.341 kg attached to the end of a thin rod
with length L0.452 m and negligible mass.
The other end of the rod is pivoted so that the
ball can move in a vertical circle. The rod is
held horizontally as shown and then given
enough of a downward push to cause the
ball to swing down and around and just reach
the vertically up position, with zero speed
there. How much work is done on the ball by
the gravitational force from the initial point
202 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
Module 8-1 Potential Energy
•1 What is the spring constant of a spring that stores 25 J of
elastic potential energy when compressed by 7.5 cm?
•2 In Fig. 8-29, a single frictionless roller-coaster car of mass
m825 kg tops the first hill with speed v017.0 m/s at height
h42.0 m. How much work does the gravitational force do on the
car from that point to (a) point A, (b) point B, and (c) point C? If the
gravitational potential energy of the carEarth system is taken to be
zero at C, what is its value when the car is at (d) Band (e) A? (f) If
mass mwere doubled,would the change in the gravitational potential
energy of the system between points Aand Bincrease, decrease, or
remain the same?
SSM
to (a) the lowest point, (b) the highest point, and (c) the point on
the right level with the initial point? If the gravitational potential
energy of the ballEarth system is taken to be zero at the initial
point, what is it when the ball reaches (d) the lowest point, (e) the
highest point, and (f) the point on the right level with the initial
point? (g) Suppose the rod were pushed harder so that the ball
passed through the highest point with a nonzero speed.Would Ug
from the lowest point to the highest point then be greater than, less
than, or the same as it was when the ball stopped at the highest
point?
•5 In Fig. 8-32, a 2.00 g ice
flake is released from the edge of a
hemispherical bowl whose radius r
is 22.0 cm. The flakebowl contact
is frictionless. (a) How much work is
done on the flake by the gravita-
tional force during the flake’s
descent to the bottom of the bowl?
(b) What is the change in the poten-
tial energy of the flakeEarth sys-
tem during that descent? (c) If that
potential energy is taken to be zero
at the bottom of the bowl, what is its
value when the flake is released? (d) If, instead, the potential en-
ergy is taken to be zero at the release point, what is its value when
the flake reaches the bottom of the bowl? (e) If the mass of the
flake were doubled, would the magnitudes of the answers to (a)
through (d) increase, decrease, or remain the same?
••6 In Fig. 8-33, a small block of
mass m0.032 kg can slide along
the frictionless loop-the-loop, with
loop radius R12 cm. The block is
released from rest at point P,at
height h5.0Rabove the bottom
of the loop. How much work does
the gravitational force do on the
block as the block travels from point
Pto (a) point Qand (b) the top of
the loop? If the gravitational poten-
tial energy of the blockEarth sys-
tem is taken to be zero at the bot-
tom of the loop, what is that potential energy when the block is (c)
at point P, (d) at point Q, and (e) at the top of the loop? (f) If, in-
stead of merely being released, the block is given some initial
speed downward along the track, do the answers to (a) through (e)
increase, decrease, or remain the same?
••7 Figure 8-34 shows a thin rod, of length L2.00 m and neg-
ligible mass, that can pivot about one end to rotate in a vertical
circle. A ball of mass m5.00 kg is attached to the other end.
The rod is pulled aside to angle u030.0and released with
initial velocity . As the ball descends to its lowest point,
(a) how much work does the gravitational force do on it and
(b) what is the change in the gravitational potential energy of
v
:
00
SSM
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
First
hill
A
B
C
hh
h/2
v0
L
r
I
ce
flake
Figure 8-32 Problems 5
and 11.
Figure 8-29 Problems 2 and 9.
Figure 8-31
Problems 4
and 14.
d
D
Figure 8-30
Problems 3 and 10.
h
P
R
Q
R
Figure 8-33 Problems 6
and 17.
••16 A 700 g block is released from rest at height h0above a ver-
tical spring with spring constant k400 N/m and negligible mass.
The block sticks to the spring and momentarily stops after com-
pressing the spring 19.0 cm. How much work is done (a) by the
block on the spring and (b) by the spring on the block? (c) What is
the value of h0? (d) If the block were released from height 2.00h0
above the spring, what would be the maximum compression of the
spring?
••17 In Problem 6, what are the magnitudes of (a) the horizontal
component and (b) the vertical component of the net force acting
on the block at point Q? (c) At what height hshould the block be
released from rest so that it is on the verge of losing contact with
the track at the top of the loop? (On the verge of losing contact
means that the normal force on the block from the track has just
then become zero.) (d) Graph the magnitude of the normal force on
the block at the top of the loop versus initial height h, for the range
h0 to h6R.
••18 (a) In Problem 7, what is the speed of the ball at the lowest
point? (b) Does the speed increase, decrease, or remain the same if
the mass is increased?
••19 Figure 8-36 shows an 8.00 kg stone
at rest on a spring.The spring is compressed
10.0 cm by the stone. (a) What is the spring
constant? (b) The stone is pushed down an
additional 30.0 cm and released.What is the
elastic potential energy of the compressed
spring just before that release? (c) What is
the change in the gravitational potential en-
ergy of the stoneEarth system when the
stone moves from the release point to its maximum height? (d) What
is that maximum height,measured from the release point?
••20 A pendulum consists of a 2.0 kg stone swinging on a
4.0 m string of negligible mass. The stone has a speed of 8.0 m/s
when it passes its lowest point. (a) What is the speed when the
string is at 60to the vertical? (b) What is the greatest angle with
the vertical that the string will reach during the stone’s motion?
(c) If the potential energy of the pendulumEarth system is taken
to be zero at the stone’s lowest point, what is the total mechanical
energy of the system?
••21 Figure 8-34 shows a pendulum of length L1.25 m. Its bob
(which effectively has all the mass) has speed v0when the cord makes
an angle u040.0with the vertical. (a) What is the speed of the bob
when it is in its lowest position if v08.00 m/s? What is the least
value that v0can have if the pendulum is to swing down and then up
(b) to a horizontal position, and (c) to a vertical position with the
cord remaining straight? (d) Do the answers to (b) and (c) increase,
decrease,or remain the same if u0is increased by a few degrees?
203
PROBLEMS
the ballEarth system? (c) If the gravita-
tional potential energy is taken to be zero
at the lowest point, what is its value just as
the ball is released? (d) Do the magnitudes
of the answers to (a) through (c) increase,
decrease, or remain the same if angle u0is
increased?
••8 A 1.50 kg snowball is fired from a cliff
12.5 m high.The snowball’s initial velocity is
14.0 m/s, directed 41.0above the horizontal.
(a) How much work is done on the snowball
by the gravitational force during its flight to
the flat ground below the cliff? (b) What is
the change in the gravitational potential en-
ergy of the snowballEarth system during
the flight? (c) If that gravitational potential
energy is taken to be zero at the height of the cliff, what is its value
when the snowball reaches the ground?
Module 8-2 Conservation of Mechanical Energy
•9 In Problem 2, what is the speed of the car at (a) point A,
(b) point B, and (c) point C? (d) How high will the car go on the
last hill, which is too high for it to cross? (e) If we substitute a sec-
ond car with twice the mass, what then are the answers to (a)
through (d)?
•10 (a) In Problem 3, what is the speed of the book when it
reaches the hands? (b) If we substituted a second book with twice
the mass, what would its speed be? (c) If, instead, the book were
thrown down, would the answer to (a) increase, decrease, or
remain the same?
•11 (a) In Problem 5, what is the speed of the flake
when it reaches the bottom of the bowl? (b) If we substituted a sec-
ond flake with twice the mass, what would its speed be? (c) If,
instead, we gave the flake an initial downward speed along the
bowl, would the answer to (a) increase, decrease, or remain the
same?
•12 (a) In Problem 8, using energy techniques rather than the
techniques of Chapter 4, find the speed of the snowball as it
reaches the ground below the cliff. What is that speed (b) if the
launch angle is changed to 41.0below the horizontal and (c) if the
mass is changed to 2.50 kg?
•13 A 5.0 g marble is fired vertically upward using a spring
gun. The spring must be compressed 8.0 cm if the marble is to just
reach a target 20 m above the marble’s position on the compressed
spring.(a) What is the change Ugin the gravitational potential en-
ergy of the marbleEarth system during the 20 m ascent?
(b) What is the change Usin the elastic potential energy of the
spring during its launch of the marble? (c) What is the spring con-
stant of the spring?
•14 (a) In Problem 4, what initial speed must be given the ball so
that it reaches the vertically upward position with zero speed? What
then is its speed at (b) the lowest point and (c) the point on the right
at which the ball is level with the initial point? (d) If the ball’s mass
were doubled, would the answers to (a) through (c) increase, de-
crease, or remain the same?
•15 In Fig. 8-35, a runaway truck with failed brakes is mov-
ing downgrade at 130 km/h just before the driver steers the truck
up a frictionless emergency escape ramp with an inclination of
u15.The truck’s mass is 1.2 104kg. (a) What minimum length
SSM
SSM
WWWSSM
Lmust the ramp have if the truck is to stop (momentarily) along
it? (Assume the truck is a particle, and justify that assumption.)
Does the minimum length Lincrease, decrease, or remain the same
if (b) the truck’s mass is decreased and (c) its speed is decreased?
L
θ
0
v0
m
Figure 8-34
Problems 7, 18,
and 21.
L
θ
Figure 8-35 Problem 15.
k
Figure 8-36
Problem 19.
Uof the ballEarth system between t0 and
t6.0 s (still free fall)?
••26 A conservative force ,
where xis in meters, acts on a particle moving
along an xaxis.The potential energy Uassociated
with this force is assigned a value of 27 J at x0.
(a) Write an expression for Uas a function of x,
with Uin joules and xin meters. (b)
What is the maximum positive poten-
tial energy? At what (c) negative
value and (d) positive value of xis the
potential energy equal to zero?
••27 Tarzan, who weighs 688 N,
swings from a cliff at the end of a vine
18 m long (Fig. 8-40). From the top of
the cliff to the bottom of the swing,he
descends by 3.2 m.The vine will break
if the force on it exceeds 950 N.
(a) Does the vine break? (b) If no,
what is the greatest force on it during
the swing? If yes, at what angle with
the vertical does it break?
F
:(6.0x12)i
ˆ N
is L120 cm long, has a ball
attached to one end, and is
fixed at its other end. The dis-
tance dfrom the fixed end to a
fixed peg at point Pis 75.0 cm.
When the initially stationary
ball is released with the string
horizontal as shown, it will
swing along the dashed arc.
What is its speed when it
reaches (a) its lowest point
••31 A block with mass m2.00 kg is placed against a spring
and (b) its highest point after
the string catches on the peg?
••24 A block of mass m2.0 kg is dropped
from height h40 cm onto a spring of spring
constant k1960 N/m (Fig. 8-39). Find the max-
imum distance the spring is compressed.
••25 At t0 a 1.0 kg ball is thrown from a tall
tower with . What isv
:(18 m/s)i
ˆ(24 m/s)j
ˆ
••23 The string in Fig. 8-38
ILW
204 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
••22 A 60 kg skier starts from rest at height H20 m above
the end of a ski-jump ramp (Fig. 8-37) and leaves the ramp at angle
u28. Neglect the effects of air resistance and assume the ramp
is frictionless. (a) What is the maximum height hof his jump above
the end of the ramp? (b) If he increased his weight by putting on a
backpack, would hthen be greater, less,or the same?
r
P
L
d
Figure 8-38 Problems 23 and 70.
H
h
θ
End of
ramp
Figure 8-37 Problem 22.
h
k
m
Figure 8-39
Problem 24.
Figure 8-40 Problem 27.
θ
Figure 8-43 Problem 30.
••28 Figure 8-41aapplies to the spring
in a cork gun (Fig. 8-41b); it shows the
spring force as a function of the stretch or
compression of the spring. The spring is
compressed by 5.5 cm and used to propel
a 3.8 g cork from the gun. (a) What is the
speed of the cork if it is released as the
spring passes through its relaxed posi-
tion? (b) Suppose, instead, that the cork
sticks to the spring and stretches it 1.5 cm
before separation occurs.What now is the
speed of the cork at the time of release?
of mass m12 kg is released from rest
on a frictionless incline of angle 30.
Below the block is a spring that can be
compressed 2.0 cm by a force of 270 N.
The block momentarily stops when
it compresses the spring by 5.5 cm.
(a) How far does the block move
down the incline from its rest posi-
tion to this stopping point? (b) What
is the speed of the block just as it
touches the spring?
••30 A 2.0 kg breadbox on a fric-
tionless incline of angle u40is
connected, by a cord that runs over a
pulley, to a light spring of spring con-
stant k120 N/m, as shown in
θ
m
Figure 8-42 Problems 29
and 35.
••29 In Fig. 8-42, a block
WWWSSM
x (cm)
2 4 –2–4
0.4
0.2
(a)
x
0
(b)
Compressed
spring
Cork
–0.2
–0.4
Force (N)
Figure 8-41 Problem 28.
Fig. 8-43. The box is released from rest when the spring is
unstretched.Assume that the pulley is massless and frictionless. (a)
What is the speed of the box when it has moved 10 cm down the in-
cline? (b) How far down the incline from its point of release does
the box slide before momentarily stopping, and what are the (c)
magnitude and (d) direction (up or down the incline) of the box’s
acceleration at the instant the box momentarily stops?
ILW
on a frictionless incline with angle 30.0(Fig. 8-44). (The block is
not attached to the spring.) The spring, with spring constant k19.6
N/cm, is compressed 20.0 cm and then released. (a) What is the elastic
potential energy of the compressed
spring? (b) What is the change in the
gravitational potential energy of the
blockEarth system as the block
moves from the release point to its
highest point on the incline? (c)
How far along the incline is the
highest point from the release
point?
θ
m
k
Figure 8-44 Problem 31.
a distance ddown a frictionless incline at angle u30.0where it
runs into a spring of spring constant 431 N/m.When the block mo-
mentarily stops, it has compressed the spring by 21.0 cm.What are
(a) distance dand (b) the distance between the point of the first
blockspring contact and the point where the block’s speed is
greatest?
•••36 Two children are
playing a game in which
they try to hit a small box
on the floor with a marble
fired from a spring-loaded
gun that is mounted on a
table.The target box is hori-
zontal distance D2.20 m
from the edge of the table;
see Fig. 8-48. Bobby compresses the spring 1.10 cm, but the center
of the marble falls 27.0 cm short of the center of the box. How far
should Rhoda compress the spring to score a direct hit? Assume
that neither the spring nor the ball encounters friction in the gun.
•••37 A uniform cord of length 25 cm and mass 15 g is initially
stuck to a ceiling. Later, it hangs vertically from the ceiling with only
one end still stuck. What is the change in the gravitational potential
energy of the cord with this change in orientation? (Hint: Consider a
differential slice of the cord and then use integral calculus.)
Module 8-3 Reading a Potential Energy Curve
••38 Figure 8-49 shows a plot of potential energy Uversus posi-
tion xof a 0.200 kg particle that can travel only along an xaxis
under the influence of a conservative force. The graph has these
••39 Figure 8-50 shows a
plot of potential energy Uver-
sus position xof a 0.90 kg parti-
cle that can travel only along an
xaxis. (Nonconservative forces
are not involved.) Three values
are
and The particle is
released at x4.5 m with an
initial speed of 7.0 m/s, headed
in the negative xdirection.
(a) If the particle can reach x1.0 m, what is its speed there, and if
it cannot, what is its turning point? What are the (b) magnitude
and (c) direction of the force on the particle as it begins to move to
the left of x4.0 m? Suppose, instead, the particle is headed in the
positive xdirection when it is released at x4.5 m at speed 7.0 m/s.
(d) If the particle can reach x7.0 m, what is its speed there, and if
it cannot, what is its turning point? What are the (e) magnitude and
(f) direction of the force on the particle as it begins to move to the
right of x5.0 m?
••40 The potential energy of a diatomic molecule (a two-atom
system like H2or O2) is given by
where ris the separation of the two atoms of the molecule and A
and Bare positive constants. This potential energy is associated
with the force that binds the two atoms together. (a) Find the equilib-
rium separation that is, the distance between the atoms at which the
force on each atom is zero. Is the force repulsive (the atoms are
pushed apart) or attractive (they are pulled together) if their separa-
tion is (b) smaller and (c) larger than the equilibrium separation?
•••41 A single conservative force F(x) acts on a 1.0 kg particle
that moves along an xaxis. The potential energy U(x) associated
with F(x) is given by
U(x)4x ex/4 J,
where xis in meters.At x5.0 m the particle has a kinetic energy
of 2.0 J. (a) What is the mechanical energy of the system? (b) Make
UA
r12 B
r6,
U
C45.0 J.
U
B35.0 J,U
A15.0 J,
k170 N/m is at the top of a fric-
tionless incline of angle 37.0.
The lower end of the incline is dis-
tance D1.00 m from the end of
the spring, which is at its relaxed
length. A 2.00 kg canister is pushed
against the spring until the spring is
compressed 0.200 m and released
from rest. (a) What is the speed of
the canister at the instant the spring
returns to its relaxed length (which is when the canister loses contact
with the spring)? (b) What is the speed of the canister when it
reaches the lower end of the incline?
•••34 A boy is initially seated
on the top of a hemispherical ice
mound of radius R13.8 m. He
begins to slide down the ice, with a
negligible initial speed (Fig. 8-47).
Approximate the ice as being fric-
tionless. At what height does the
boy lose contact with the ice?
•••35 In Fig. 8-42, a block of mass m3.20 kg slides from rest
••32 In Fig. 8-45, a chain is held
on a frictionless table with one-
fourth of its length hanging over
the edge. If the chain has length
L28 cm and mass m0.012 kg,
how much work is required to pull
the hanging part back onto the
table?
•••33 In Fig. 8-46, a spring with
205
PROBLEMS
values: , and .The particle is
released at the point where Uforms a “potential hill” of “height”
, with kinetic energy 4.00 J. What is the speed of the
particle at (a) m and (b) m? What is the position
of the turning point on (c) the right side and (d) the left side?
x6.5x3.5
U
B12.00 J
U
D24.00 JU
A9.00 J, U
C20.00 J
D
θ
Figure 8-46 Problem 33.
D
Figure 8-48 Problem 36.
R
Figure 8-47 Problem 34.
Figure 8-45 Problem 32.
UC
UB
UA
246
x (m)
U (J)
Figure 8-50 Problem 39.
UA
UB
UC
UD
U (J)
1 2 3 4 5 6 7 8 9 0
x (m)
Figure 8-49 Problem 38.
•50 A 60 kg skier leaves the end of a ski-jump ramp with a ve-
locity of 24 m/s directed 25above the horizontal. Suppose that as a
result of air drag the skier returns to the ground with a speed of 22
m/s, landing 14 m vertically below the end of the ramp. From the
launch to the return to the ground, by how much is the mechanical
energy of the skierEarth system reduced because of air drag?
•51 During a rockslide, a 520 kg rock slides from rest down a hill-
side that is 500 m long and 300 m high. The coefficient of kinetic
friction between the rock and the hill surface is 0.25. (a) If the grav-
itational potential energy Uof the rockEarth system is zero at
the bottom of the hill, what is the value of Ujust before the slide?
(b) How much energy is transferred to thermal energy during the
slide? (c) What is the kinetic energy of the rock as it reaches the
bottom of the hill? (d) What is its speed then?
••52 A large fake cookie sliding on a horizontal surface is
attached to one end of a horizontal spring with spring constant
k400 N/m; the other end of the spring is fixed in place. The
cookie has a kinetic energy of 20.0 J as it passes through
the spring’s equilibrium position. As the cookie slides, a frictional
force of magnitude 10.0 N acts on it. (a) How far will the cookie
slide from the equilibrium position before coming momentarily to
rest? (b) What will be the kinetic energy of the cookie as it slides
back through the equilibrium position?
••53 In Fig. 8-52, a 3.5 kg
block is accelerated from rest
by a compressed spring of
spring constant 640 N/m. The
block leaves the spring at the
spring’s relaxed length and
then travels over a horizontal
floor with a coefficient of ki-
netic friction mk0.25. The frictional force stops the block in dis-
tance D7.8 m. What are (a) the increase in the thermal energy
of the blockfloor system, (b) the maximum kinetic energy of the
block, and (c) the original compression distance of the spring?
••54 A child whose weight is 267 N slides down a 6.1 m play-
ground slide that makes an angle of 20with the horizontal.The co-
efficient of kinetic friction between slide and child is 0.10. (a) How
much energy is transferred to thermal energy? (b) If she starts at
the top with a speed of 0.457 m/s, what is her speed at the bottom?
••55 In Fig. 8-53, a block of mass m2.5 kg slides head on
into a spring of spring constant
k320 N/m. When the block
ILW
stops, it has compressed the
spring by 7.5 cm. The coefficient
of kinetic friction between block
and floor is 0.25. While the block
is in contact with the spring and
being brought to rest, what are (a)
the work done by the spring force and (b) the increase in thermal
energy of the blockfloor system? (c) What is the block’s speed
just as it reaches the spring?
••56 You push a 2.0 kg block against a horizontal spring, com-
pressing the spring by 15 cm. Then you release the block, and the
206 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
a plot of U(x) as a function of xfor 0 x10 m, and on the same
graph draw the line that represents the mechanical energy of the
system. Use part (b) to determine (c) the least value of xthe parti-
cle can reach and (d) the greatest value of xthe particle can reach.
Use part (b) to determine (e) the maximum kinetic energy of the
particle and (f) the value of xat which it occurs. (g) Determine an
expression in newtons and meters for F(x) as a function of x. (h)
For what (finite) value of xdoes F(x)0?
Module 8-4 Work Done on a System by an External Force
•42 A worker pushed a 27 kg block 9.2 m along a level floor at con-
stant speed with a force directed 32below the horizontal. If the coef-
ficient of kinetic friction between block and floor was 0.20, what were
(a) the work done by the worker’s force and (b) the increase in ther-
mal energy of the blockfloor system?
•43 A collie drags its bed box across a floor by applying a hori-
zontal force of 8.0 N. The kinetic frictional force acting on the box
has magnitude 5.0 N. As the box is dragged through 0.70 m along
the way, what are (a) the work done by the collie’s applied force
and (b) the increase in thermal energy of the bed and floor?
••44 A horizontal force of magnitude 35.0 N pushes a block of
mass 4.00 kg across a floor where the coefficient of kinetic friction is
0.600. (a) How much work is done by that applied force on the
blockfloor system when the block slides through a displacement of
3.00 m across the floor? (b) During that displacement, the thermal
energy of the block increases by 40.0 J. What is the increase in ther-
mal energy of the floor? (c) What is the increase in the kinetic energy
of the block?
••45 A rope is used to pull a 3.57 kg block at constant speed
4.06 m along a horizontal floor. The force on the block from the
rope is 7.68 N and directed 15.0above the horizontal.What are (a)
the work done by the rope’s force, (b) the increase in thermal en-
ergy of the blockfloor system, and (c) the coefficient of kinetic
friction between the block and floor?
Module 8-5 Conservation of Energy
•46 An outfielder throws a baseball with an initial speed of
81.8 mi/h. Just before an infielder catches the ball at the same
level, the ball’s speed is 110 ft/s. In foot-pounds, by how much is the
mechanical energy of the ballEarth system reduced because of
air drag? (The weight of a baseball is 9.0 oz.)
•47 A 75 g Frisbee is thrown from a point 1.1 m above the
ground with a speed of 12 m/s. When it has reached a height of
2.1 m, its speed is 10.5 m/s. What was the reduction in Emec of the
FrisbeeEarth system because of air drag?
•48 In Fig. 8-51, a block slides
down an incline. As it moves from
point Ato point B, which are 5.0 m
apart, force acts on the block, with
magnitude 2.0 N and directed down
the incline.The magnitude of the fric-
tional force acting on the block is
10 N. If the kinetic energy of the
block increases by 35 J between Aand B, how much work is done on
the block by the gravitational force as the block moves from Ato B?
•49 A 25 kg bear slides, from rest, 12 m down a
lodgepole pine tree, moving with a speed of 5.6 m/s just before
hitting the ground. (a) What change occurs in the gravitational
ILWSSM
F
:
SSM
potential energy of the bearEarth system during the slide?
(b) What is the kinetic energy of the bear just before hitting the
ground? (c) What is the average frictional force that acts on the
sliding bear?
x
0
Figure 8-53 Problem 55.
No friction D
(k)
μ
Figure 8-52 Problem 53.
A
B
Figure 8-51 Problems 48
and 71.
•••65 A particle can slide along a
track with elevated ends and a flat
central part, as shown in Fig. 8-58.
The flat part has length L40 cm.
The curved portions of the track are
frictionless, but for the flat part the
coefficient of kinetic friction is mk
0.20.The particle is released from rest at point A, which is at height
hL/2. How far from the left edge of the flat part does the particle
finally stop?
Additional Problems
66 A 3.2 kg sloth hangs 3.0 m above the ground. (a) What is the
gravitational potential energy of the slothEarth system if we take
the reference point y0 to be at the ground? If the sloth drops
to the ground and air drag on it is assumed to be negligible, what
are the (b) kinetic energy and (c) speed of the sloth just before it
reaches the ground?
207
PROBLEMS
spring sends it sliding across a tabletop. It stops 75 cm from where
you released it. The spring constant is 200 N/m. What is the
blocktable coefficient of kinetic friction?
••57 In Fig. 8-54, a block slides along a track from one level to a
higher level after passing through an intermediate valley. The track
is frictionless until the block reaches the higher level. There a fric-
tional force stops the block in a distance d. The block’s initial speed
v0is 6.0 m/s,the height difference his 1.1 m, and mkis 0.60.Find d.
•••63 The cable of the 1800 kg elevator
cab in Fig. 8-56 snaps when the cab is at
rest at the first floor, where the cab bottom
is a distance d3.7 m above a spring of
spring constant k0.15 MN/m. A safety
device clamps the cab against guide rails so
that a constant frictional force of 4.4 kN
opposes the cab’s motion. (a) Find the
speed of the cab just before it hits the
spring. (b) Find the maximum distance x
that the spring is compressed (the fric-
tional force still acts during this compres-
sion). (c) Find the distance that the cab
will bounce back up the shaft. (d) Using
conservation of energy, find the approxi-
mate total distance that the cab will move before coming to rest.
(Assume that the frictional force on the cab is negligible when the
cab is stationary.)
•••64 In Fig. 8-57, a block is released from rest at height d40
cm and slides down a frictionless ramp and onto a first plateau,
which has length dand where the coefficient of kinetic friction is
0.50. If the block is still moving, it then slides down a second fric-
tionless ramp through height d/2 and onto a lower plateau, which
has length d/2 and where the coefficient of kinetic friction is
again 0.50. If the block is still moving, it then slides up a friction-
less ramp until it (momentarily) stops. Where does the block
stop? If its final stop is on a plateau, state which one and give the
distance Lfrom the left edge of that plateau. If the block reaches
the ramp, give the height Habove the lower plateau where it
momentarily stops.
h
μ
k
μ
= 0
v0
d
Figure 8-54 Problem 57.
••58 A cookie jar is moving up a 40incline. At a point 55 cm
from the bottom of the incline (measured along the incline), the jar
has a speed of 1.4 m/s. The coefficient of kinetic friction between
jar and incline is 0.15. (a) How much farther up the incline will the
jar move? (b) How fast will it be going when it has slid back to the
bottom of the incline? (c) Do the answers to (a) and (b) increase,
decrease, or remain the same if we decrease the coefficient of ki-
netic friction (but do not change the given speed or location)?
••59 A stone with a weight of 5.29 N is launched vertically from
ground level with an initial speed of 20.0 m/s, and the air drag on it
is 0.265 N throughout the flight. What are (a) the maximum height
reached by the stone and (b) its speed just before it hits the ground?
••60 A 4.0 kg bundle starts up a 30incline with 128 J of kinetic
energy. How far will it slide up the incline if the coefficient of ki-
netic friction between bundle and incline is 0.30?
••61 When a click beetle is upside down on its back, it jumps
upward by suddenly arching its back, transferring energy stored in a
muscle to mechanical energy.This launching mechanism produces an
audible click, giving the beetle its name.Videotape of a certain click-
beetle jump shows that a beetle of mass m4.0 106kg moved di-
rectly upward by 0.77 mm during the launch and then to a maximum
height of h0.30 m. During the launch, what are the average mag-
nitudes of (a) the external force on the beetle’s back from the floor
and (b) the acceleration of the beetle in terms of g?
•••62 In Fig. 8-55, a block slides along a path that is without fric-
tion until the block reaches the section of length L0.75 m, which
begins at height h2.0 m on a ramp of angle u30. In that sec-
tion, the coefficient of kinetic friction is 0.40. The block passes
through point Awith a speed of 8.0 m/s. If the block can reach point
B(where the friction ends), what is its speed there, and if it cannot,
what is its greatest height above A?
θ
A
h
L
B
Figure 8-55 Problem 62.
d
d/2
d
d/2
Figure 8-57 Problem 64.
L
h
A
Figure 8-58 Problem 65.
d
k
Figure 8-56
Problem 63.
between snow and skis would make him stop just at the top of
the lower peak?
73 The temperature of a plastic cube is monitored while the
cube is pushed 3.0 m across a floor at constant speed by a horizon-
tal force of 15 N. The thermal energy of the cube increases by 20 J.
What is the increase in the thermal energy of the floor along which
the cube slides?
74 A skier weighing 600 N goes over a frictionless circular hill
of radius R20 m (Fig. 8-62). Assume that the effects of air re-
sistance on the skier are negligible. As she comes up the hill, her
speed is 8.0 m/s at point B, at angle u20. (a) What is her
speed at the hilltop (point A) if she coasts without using her
poles? (b) What minimum speed can she have at Band still coast
to the hilltop? (c) Do the answers to these two questions in-
crease, decrease, or remain the same if the skier weighs 700 N
instead of 600 N?
SSM
208 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
67 A spring (k200 N/m) is
fixed at the top of a frictionless plane
inclined at angle 40(Fig. 8-59). A
1.0 kg block is projected up the plane,
from an initial position that is distance
d0.60 m from the end of the relaxed
spring, with an initial kinetic energy of
16 J. (a) What is the kinetic energy of
the block at the instant it has com-
pressed the spring 0.20 m? (b) With
what kinetic energy must the block be
projected up the plane if it is to stop momentarily when it has
compressed the spring by 0.40 m?
68 From the edge of a cliff, a 0.55 kg projectile is launched
with an initial kinetic energy of 1550 J.The projectile’s maximum
upward displacement from the launch point is 140 m. What
are the (a) horizontal and (b) vertical components of its launch
velocity? (c) At the instant the vertical component of its velocity
is 65 m/s, what is its vertical displacement from the launch
point?
69 In Fig. 8-60, the pulley has
negligible mass, and both it and the
inclined plane are frictionless. Block
Ahas a mass of 1.0 kg, block Bhas a
mass of 2.0 kg, and angle uis 30.If
the blocks are released from rest
with the connecting cord taut, what
is their total kinetic energy when
block Bhas fallen 25 cm?
70 In Fig. 8-38, the string is L120 cm long, has a ball
attached to one end, and is fixed at its other end. A fixed peg is at
point P. Released from rest, the ball swings down until the string
catches on the peg; then the ball swings up, around the peg. If the
ball is to swing completely around the peg, what value must dis-
tance dexceed? (Hint: The ball must still be moving at the top of
its swing.Do you see why?)
71 In Fig. 8-51, a block is sent sliding down a frictionless
ramp. Its speeds at points Aand Bare 2.00 m/s and 2.60 m/s, re-
spectively. Next, it is again sent sliding down the ramp, but this
time its speed at point Ais 4.00 m/s. What then is its speed at
point B?
72 Two snowy peaks are at heights H850 m and h750 m
above the valley between them. A ski run extends between
the peaks, with a total length of 3.2 km and an average slope of
30(Fig. 8-61). (a) A skier starts from rest at the top of
the higher peak. At what speed will he arrive at the top of
the lower peak if he coasts without using ski poles? Ignore fric-
tion. (b) Approximately what coefficient of kinetic friction
SSM
SSM
SSM
d
θ
Figure 8-59 Problem 67.
Outward Inward
(a) 3.0 3.0
(b) 5.0 5.0
(c) 2.0x2.0x
(d) 3.0x23.0x2
θ
B
A
Figure 8-60 Problem 69.
H
h
θ θ
Figure 8-61 Problem 72.
R
θ
A
B
Figure 8-62 Problem 74.
Find the net work done on the particle by the external force for the
round trip for each of the four situations.(e) For which, if any, is the
external force conservative?
77 A conservative force F(x) acts on a 2.0 kg particle that
moves along an xaxis. The potential energy U(x) associated with
F(x) is graphed in Fig. 8-63. When the particle is at x2.0 m, its
SSM
75 To form a pendulum, a 0.092 kg ball is attached to one end
of a rod of length 0.62 m and negligible mass, and the other end of
the rod is mounted on a pivot. The rod is rotated until it is straight
up, and then it is released from rest so that it swings down around
the pivot. When the ball reaches its lowest point, what are (a) its
speed and (b) the tension in the rod? Next, the rod is rotated until it
is horizontal, and then it is again released from rest. (c) At what an-
gle from the vertical does the tension in the rod equal the weight of
the ball? (d) If the mass of the ball is increased, does the answer to
(c) increase, decrease, or remain the same?
76 We move a particle along an xaxis, first outward from x1.0 m
to x4.0 m and then back to x1.0 m, while an external force
acts on it. That force is directed along the xaxis, and its xcompo-
nent can have different values for the outward trip and for the re-
turn trip. Here are the values (in newtons) for four situations, where
xis in meters:
SSM
209
PROBLEMS
velocity is 1.5 m/s. What are the (a) magnitude and (b) direction
of F(x) at this position? Between what positions on the (c) left and
(d) right does the particle move? (e) What is the particle’s speed at
x7.0 m?
81 A particle can move along only an xaxis, where conservative
forces act on it (Fig. 8-66 and the following table). The particle is
released at x5.00 m with a kinetic energy of K14.0 J and a
potential energy of U0. If its motion is in the negative direction
of the xaxis, what are its (a) Kand (b) Uat x2.00 m and its
(c) Kand (d) Uat x0? If its motion is in the positive direction of
the xaxis, what are its (e) Kand (f) Uat x11.0 m, its (g) Kand
(h) Uat x12.0 m, and its (i) Kand ( j) Uat x13.0 m? (k) Plot
U(x) versus xfor the range x0 to x13.0 m.
0
–5
–10
–15
–20
U(x) ( J)
5
x(m)
10 150
Figure 8-63 Problem 77.
78 At a certain factory, 300 kg
crates are dropped vertically from
a packing machine onto a conveyor
belt moving at 1.20 m/s (Fig. 8-64).
(A motor maintains the belt’s con-
stant speed.) The coefficient of ki-
netic friction between the belt and
each crate is 0.400. After a short
time, slipping between the belt and
the crate ceases, and the crate then moves along with the belt. For
the period of time during which the crate is being brought to rest
relative to the belt, calculate, for a coordinate system at rest in
the factory, (a) the kinetic energy supplied to the crate, (b) the
magnitude of the kinetic frictional force acting on the crate, and
(c) the energy supplied by the motor. (d) Explain why answers
(a) and (c) differ.
79 A 1500 kg car begins sliding down a 5.0inclined road
with a speed of 30 km/h. The engine is turned off, and the only
forces acting on the car are a net frictional force from the road and
the gravitational force. After the car has traveled 50 m along the
road, its speed is 40 km/h. (a) How much is the mechanical energy
of the car reduced because of the net frictional force? (b) What is
the magnitude of that net frictional force?
80 In Fig. 8-65, a 1400 kg block of granite is pulled up an incline
at a constant speed of 1.34 m/s by a cable and winch.The indicated
distances are d140 m and d230 m. The coefficient of kinetic
friction between the block and the incline is 0.40. What is the
power due to the force applied to the block by the cable?
SSM
d2
d1
Figure 8-65 Problem 80.
FRAGILE
FRAGILE
Figure 8-64 Problem 78.
0 2 4 6 8 10 12
F1F2F3F4
x (m)
Figure 8-66 Problems 81 and 82.
Range Force
0 to 2.00 m
2.00 m to 3.00 m
3.00 m to 8.00 m F0
8.00 m to 11.0 m
11.0 m to 12.0 m
12.0 m to 15.0 m F0
F
:
4(1.00 N)i
ˆ
F
:
3(4.00 N)i
ˆ
F
:
2(5.00 N)i
ˆ
F
:
1(3.00 N)i
ˆ
Next, the particle is released from rest at x0. What are (l) its
kinetic energy at x5.0 m and (m) the maximum positive position
xmax it reaches? (n) What does the particle do after it reaches xmax?
82 For the arrangement of forces in Problem 81, a 2.00 kg parti-
cle is released at x5.00 m with an initial velocity of 3.45 m/s in
the negative direction of the xaxis. (a) If the particle can reach
x0 m, what is its speed there, and if it cannot, what is its turning
point? Suppose, instead, the particle is headed in the positive xdi-
rection when it is released at x5.00 m at speed 3.45 m/s. (b) If
the particle can reach x13.0 m, what is its speed there, and if it
cannot, what is its turning point?
83 A 15 kg block is accelerated at 2.0 m/s2along a horizon-
tal frictionless surface, with the speed increasing from 10 m/s to
30 m/s. What are (a) the change in the block’s mechanical energy
and (b) the average rate at which energy is transferred to the
block? What is the instantaneous rate of that transfer when the
block’s speed is (c) 10 m/s and (d) 30 m/s?
84 A certain spring is found not to conform to Hooke’s law. The
force (in newtons) it exerts when stretched a distance x(in meters)
is found to have magnitude 52.8x38.4x2in the direction oppos-
ing the stretch. (a) Compute the work required to stretch the
spring from x0.500 m to x1.00 m. (b) With one end of the
spring fixed, a particle of mass 2.17 kg is attached to the other end
of the spring when it is stretched by an amount x1.00 m. If the
particle is then released from rest, what is its speed at the instant
the stretch in the spring is x0.500 m? (c) Is the force exerted by
the spring conservative or nonconservative? Explain.
85 Each second, 1200 m3of water passes over a waterfall
100 m high. Three-fourths of the kinetic energy gained by the water
in falling is transferred to electrical energy by a hydroelectric gener-
ator. At what rate does the generator produce electrical energy?
(The mass of 1 m3of water is 1000 kg.)
SSM
SSM
a spring of spring constant k200 N/m that has one end fixed, as
shown in Fig. 8-69. The horizontal
surface and the pulley are friction-
less, and the pulley has negligible
mass. The blocks are released from
rest with the spring relaxed.
(a) What is the combined kinetic
energy of the two blocks when the
hanging block has fallen 0.090 m?
(b) What is the kinetic energy of
the hanging block when it has
fallen that 0.090 m? (c) What maxi-
mum distance does the hanging block fall before momentarily
stopping?
92 A volcanic ash flow is moving across horizontal ground when
it encounters a 10upslope. The front of the flow then travels 920
m up the slope before stopping. Assume that the gases entrapped
in the flow lift the flow and thus make the frictional force from the
ground negligible; assume also that the mechanical energy of
the front of the flow is conserved.What was the initial speed of the
front of the flow?
93 A playground slide is in the form of an arc of a circle that has
a radius of 12 m.The maximum height of the slide is h4.0 m, and
the ground is tangent to the circle (Fig. 8-70). A 25 kg child starts
from rest at the top of the slide and has a speed of 6.2 m/s at the
bottom. (a) What is the length of the slide? (b) What average fric-
tional force acts on the child over this distance? If, instead of the
ground, a vertical line through the top of the slide is tangent to the
circle, what are (c) the length of the slide and (d) the average fric-
tional force on the child?
210 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
86 In Fig. 8-67, a small block is sent through point Awith a
speed of 7.0 m/s. Its path is without friction until it reaches the sec-
tion of length L12 m, where the coefficient of kinetic friction is
0.70. The indicated heights are h16.0 m and h22.0 m. What
are the speeds of the block at (a) point Band (b) point C? (c) Does
the block reach point D? If so, what is its speed there; if not, how far
through the section of friction does it travel?
91 Two blocks, of masses M2.0 kg and 2M, are connected to
A
B
C D
h1
h2L
Figure 8-67 Problem 86.
87 A massless rigid rod of
length Lhas a ball of mass m
attached to one end (Fig. 8-68). The
other end is pivoted in such a way
that the ball will move in a vertical
circle. First, assume that there is no
friction at the pivot. The system is
launched downward from the hori-
zontal position Awith initial speed
v0. The ball just barely reaches point
Dand then stops. (a) Derive an ex-
pression for v0in terms of L,m, and
g. (b) What is the tension in the rod
when the ball passes through B? (c) A little grit is placed on the
pivot to increase the friction there. Then the ball just barely
reaches Cwhen launched from Awith the same speed as before.
What is the decrease in the mechanical energy during this motion?
(d) What is the decrease in the mechanical energy by the time the
ball finally comes to rest at Bafter several oscillations?
88 A 1.50 kg water balloon is shot straight up with an initial speed
of 3.00 m/s. (a) What is the kinetic energy of the balloon just as it is
launched? (b) How much work does the gravitational force do on
the balloon during the balloon’s full ascent? (c) What is the change
in the gravitational potential energy of the balloonEarth system
during the full ascent? (d) If the gravitational potential energy is
taken to be zero at the launch point, what is its value when the bal-
loon reaches its maximum height? (e) If, instead, the gravitational
potential energy is taken to be zero at the maximum height, what is
its value at the launch point? (f) What is the maximum height?
89 A 2.50 kg beverage can is thrown directly downward from a
height of 4.00 m, with an initial speed of 3.00 m/s. The air drag on
the can is negligible. What is the kinetic energy of the can (a) as it
reaches the ground at the end of its fall and (b) when it is halfway
to the ground? What are (c) the kinetic energy of the can and (d)
the gravitational potential energy of the canEarth system 0.200 s
before the can reaches the ground? For the latter, take the refer-
ence point y0 to be at the ground.
90 A constant horizontal force moves a 50 kg trunk 6.0 m up a
30incline at constant speed. The coefficient of kinetic friction is
0.20. What are (a) the work done by the applied force and (b) the
increase in the thermal energy of the trunk and incline?
SSM
2M
M
Figure 8-69 Problem 91.
h
Figure 8-70 Problem 93.
v0
D
B
ALC
Pivot
point
Rod
Figure 8-68 Problem 87.
94 The luxury liner Queen Elizabeth 2 has a diesel-electric
power plant with a maximum power of 92 MW at a cruising speed
of 32.5 knots. What forward force is exerted on the ship at this
speed? (1 knot 1.852 km/h.)
95 A factory worker accidentally releases a 180 kg crate that was
being held at rest at the top of a ramp that is 3.7 m long and in-
clined at 39to the horizontal.The coefficient of kinetic friction be-
tween the crate and the ramp, and between the crate and the hori-
zontal factory floor, is 0.28. (a) How fast is the crate moving as it
reaches the bottom of the ramp? (b) How far will it subsequently
slide across the floor? (Assume that the crate’s kinetic energy does
not change as it moves from the ramp onto the floor.) (c) Do the
answers to (a) and (b) increase, decrease, or remain the same if we
halve the mass of the crate?
96 If a 70 kg baseball player steals home by sliding into the plate
with an initial speed of 10 m/s just as he hits the ground, (a) what
211
PROBLEMS
is the decrease in the player’s kinetic energy and (b) what is the
increase in the thermal energy of his body and the ground along
which he slides?
97 A 0.50 kg banana is thrown directly upward with an initial
speed of 4.00 m/s and reaches a maximum height of 0.80 m. What
change does air drag cause in the mechanical energy of the
bananaEarth system during the ascent?
98 A metal tool is sharpened by being held against the rim of a
wheel on a grinding machine by a force of 180 N. The frictional
forces between the rim and the tool grind off small pieces of the
tool. The wheel has a radius of 20.0 cm and rotates at 2.50 rev/s.
The coefficient of kinetic friction between the wheel and the tool is
0.320. At what rate is energy being transferred from the motor
driving the wheel to the thermal energy of the wheel and tool and
to the kinetic energy of the material thrown from the tool?
99 A swimmer moves through the water at an average speed of
0.22 m/s. The average drag force is 110 N. What average power is
required of the swimmer?
100 An automobile with passengers has weight 16 400 N and is
moving at 113 km/h when the driver brakes, sliding to a stop. The
frictional force on the wheels from the road has a magnitude of
8230 N. Find the stopping distance.
101 A 0.63 kg ball thrown directly upward with an initial speed
of 14 m/s reaches a maximum height of 8.1 m. What is the change
in the mechanical energy of the ballEarth system during the
ascent of the ball to that maximum height?
102 The summit of Mount Everest is 8850 m above sea level.
(a) How much energy would a 90 kg climber expend against the
gravitational force on him in climbing to the summit from sea
level? (b) How many candy bars, at 1.25 MJ per bar, would supply
an energy equivalent to this? Your answer should suggest that
work done against the gravitational force is a very small part of the
energy expended in climbing a mountain.
103 A sprinter who weighs 670 N runs the first 7.0 m of a race in
1.6 s, starting from rest and accelerating uniformly. What are the
sprinter’s (a) speed and (b) kinetic energy at the end of the 1.6 s?
(c) What average power does the sprinter generate during the 1.6 s
interval?
104 A 20 kg object is acted on by a conservative force given by
F3.0x5.0x2, with Fin newtons and xin meters. Take the
potential energy associated with the force to be zero when the
object is at x0. (a) What is the potential energy of the system
associated with the force when the object is at x2.0 m? (b) If
the object has a velocity of 4.0 m/s in the negative direction of the
xaxis when it is at x5.0 m, what is its speed when it passes
through the origin? (c) What are the answers to (a) and (b) if the
potential energy of the system is taken to be 8.0 J when the ob-
ject is at x0?
105 A machine pulls a 40 kg trunk 2.0 m up a 40ramp at con-
stant velocity, with the machine’s force on the trunk directed paral-
lel to the ramp. The coefficient of kinetic friction between the
trunk and the ramp is 0.40. What are (a) the work done on the
trunk by the machine’s force and (b) the increase in thermal en-
ergy of the trunk and the ramp?
106 The spring in the muzzle of a child’s spring gun has a spring
constant of 700 N/m.To shoot a ball from the gun, first the spring is
compressed and then the ball is placed on it. The gun’s trigger then
releases the spring, which pushes the ball through the muzzle. The
ball leaves the spring just as it leaves the outer end of the muzzle.
When the gun is inclined upward by 30to the horizontal,a 57 g ball
is shot to a maximum height of 1.83 m above the gun’s muzzle.
Assume air drag on the ball is negligible. (a) At what speed does
the spring launch the ball? (b) Assuming that friction on the ball
within the gun can be neglected, find the spring’s initial compres-
sion distance.
107 The only force acting on a particle is conservative force . If
the particle is at point A, the potential energy of the system associ-
ated with and the particle is 40 J. If the particle moves from point
Ato point B, the work done on the particle by is 25 J. What is
the potential energy of the system with the particle at B?
108 In 1981, Daniel Goodwin climbed 443 m up the exterior of
the Sears Building in Chicago using suction cups and metal clips.
(a) Approximate his mass and then compute how much energy he
had to transfer from biomechanical (internal) energy to the gravi-
tational potential energy of the EarthGoodwin system to lift
himself to that height. (b) How much energy would he have had to
transfer if he had, instead, taken the stairs inside the building (to
the same height)?
109 A 60.0 kg circus performer slides 4.00 m down a pole to the
circus floor, starting from rest. What is the kinetic energy of the
performer as she reaches the floor if the frictional force on her
from the pole (a) is negligible (she will be hurt) and (b) has a mag-
nitude of 500 N?
110 A 5.0 kg block is projected at 5.0 m/s up a plane that is
inclined at 30with the horizontal. How far up along the
plane does the block go (a) if the plane is frictionless and (b) if the
coefficient of kinetic friction between the block and the plane is
0.40? (c) In the latter case, what is the increase in thermal energy
of block and plane during the block’s ascent? (d) If the block then
slides back down against the frictional force, what is the block’s
speed when it reaches the original projection point?
111 A 9.40 kg projectile is fired vertically upward. Air drag de-
creases the mechanical energy of the projectile–Earth system by
68.0 kJ during the projectile’s ascent. How much higher would the
projectile have gone were air drag negligible?
112 A 70.0 kg man jumping from a window lands in an elevated
fire rescue net 11.0 m below the window. He momentarily stops
when he has stretched the net by 1.50 m. Assuming that mechani-
cal energy is conserved during this process and that the net func-
tions like an ideal spring,find the elastic potential energy of the net
when it is stretched by 1.50 m.
113 A 30 g bullet moving a horizontal velocity of 500 m/s comes
to a stop 12 cm within a solid wall. (a) What is the change in the
bullet’s mechanical energy? (b) What is the magnitude of the aver-
age force from the wall stopping it?
114 A 1500 kg car starts from rest on a horizontal road and
gains a speed of 72 km/h in 30 s. (a) What is its kinetic energy at
the end of the 30 s? (b) What is the average power required of the
car during the 30 s interval? (c) What is the instantaneous power
at the end of the 30 s interval, assuming that the acceleration is
constant?
115 A 1.50 kg snowball is shot upward at an angle of 34.0to the
horizontal with an initial speed of 20.0 m/s. (a) What is its initial
kinetic energy? (b) By how much does the gravitational potential
F
:
F
:
F
:
124 The magnitude of the gravitational force between a particle
of mass m1and one of mass m2is given by
F(x)
where Gis a constant and xis the distance between the particles.
(a) What is the corresponding potential energy function U(x)?
Assume that U(x) 0 as xand that xis positive. (b) How
much work is required to increase the separation of the particles
from xx1to xx1d?
125 Approximately 5.5 106kg of water falls 50 m over
Niagara Falls each second. (a) What is the decrease in the gravi-
tational potential energy of the water–Earth system each sec-
ond? (b) If all this energy could be converted to electrical
energy (it cannot be), at what rate would electrical energy be
supplied? (The mass of 1 m3of water is 1000 kg.) (c) If the elec-
trical energy were sold at 1 cent/kW h, what would be the yearly
income?
126 To make a pendulum, a 300 g ball is attached to one end of
a string that has a length of 1.4 m and negligible mass. (The other
end of the string is fixed.) The ball is pulled to one side until the
string makes an angle of 30.0with the vertical; then (with
the string taut) the ball is released from rest. Find (a) the speed of
the ball when the string makes an angle of 20.0with the vertical
and (b) the maximum speed of the ball. (c) What is the angle be-
tween the string and the vertical when the speed of the ball is
one-third its maximum value?
127 In a circus act, a 60 kg clown is shot from a cannon with an
initial velocity of 16 m/s at some unknown angle above the hori-
zontal.A short time later the clown lands in a net that is 3.9 m ver-
tically above the clown’s initial position. Disregard air drag. What
is the kinetic energy of the clown as he lands in the net?
128 A 70 kg firefighter slides, from rest, 4.3 m down a vertical
pole. (a) If the firefighter holds onto the pole lightly, so that the
frictional force of the pole on her is negligible, what is her speed
just before reaching the ground floor? (b) If the firefighter grasps
the pole more firmly as she slides, so that the average frictional
force of the pole on her is 500 N upward, what is her speed just be-
fore reaching the ground floor?
129 The surface of the continental United States has an area of
about 8 106km2and an average elevation of about 500 m
(above sea level). The average yearly rainfall is 75 cm. The frac-
tion of this rainwater that returns to the atmosphere by evapora-
tion is ; the rest eventually flows into the ocean. If the decrease
in gravitational potential energy of the water–Earth system asso-
ciated with that flow could be fully converted to electrical en-
ergy, what would be the average power? (The mass of 1 m3of
water is 1000 kg.)
130 A spring with spring constant k200 N/m is suspended
vertically with its upper end fixed to the ceiling and its lower
end at position y0. A block of weight 20 N is attached to the
lower end, held still for a moment, and then released. What are
(a) the kinetic energy K, (b) the change (from the initial value)
in the gravitational potential energy Ug, and (c) the change in
the elastic potential energy Ueof the spring–block system when
the block is at y5.0 cm? What are (d) K,(e) Ug, and (f) Ue
when y10 cm, (g) K, (h) Ug, and (i) Uewhen y15 cm,
and (j) K, (k) Ug, and (l) Uewhen y 20 cm?
2
3
::
Gm1m2
x2,
energy of the snowballEarth system change as the snowball
moves from the launch point to the point of maximum height? (c)
What is that maximum height?
116 A 68 kg sky diver falls at a constant terminal speed of
59 m/s. (a) At what rate is the gravitational potential energy of the
Earthsky diver system being reduced? (b) At what rate is the sys-
tem’s mechanical energy being reduced?
117 A 20 kg block on a horizontal surface is attached to a hori-
zontal spring of spring constant k4.0 kN/m. The block is pulled
to the right so that the spring is stretched 10 cm beyond its relaxed
length, and the block is then released from rest.The frictional force
between the sliding block and the surface has a magnitude of 80 N.
(a) What is the kinetic energy of the block when it has moved
2.0 cm from its point of release? (b) What is the kinetic energy of
the block when it first slides back through the point at which the
spring is relaxed? (c) What is the maximum kinetic energy attained
by the block as it slides from its point of release to the point at
which the spring is relaxed?
118 Resistance to the motion of an automobile consists of road
friction, which is almost independent of speed, and air drag, which
is proportional to speed-squared. For a certain car with a weight of
12 000 N, the total resistant force Fis given by F300 1.8v2,
with Fin newtons and vin meters per second. Calculate the power
212 CHAPTER 8 POTENTIAL ENERGY AND CONSERVATION OF ENERGY
(in horsepower) required to accelerate the car at 0.92 m/s2when
the speed is 80 km/h.
119 A 50 g ball is thrown from a window with an initial
SSM
velocity of 8.0 m/s at an angle of 30above the horizontal. Using
energy methods, determine (a) the kinetic energy of the ball at the
top of its flight and (b) its speed when it is 3.0 m below the window.
Does the answer to (b) depend on either (c) the mass of the ball or
(d) the initial angle?
120 A spring with a spring constant of 3200 N/m is initially
stretched until the elastic potential energy of the spring is 1.44 J.
(U0 for the relaxed spring.) What is Uif the initial stretch is
changed to (a) a stretch of 2.0 cm, (b) a compression of 2.0 cm, and
(c) a compression of 4.0 cm?
121 A locomotive with a power capability of 1.5 MW can
accelerate a train from a speed of 10 m/s to 25 m/s in 6.0 min. (a)
Calculate the mass of the train. Find (b) the speed of the train and
(c) the force accelerating the train as functions of time (in seconds)
during the 6.0 min interval. (d) Find the distance moved by the
train during the interval.
122 A 0.42 kg shuffleboard disk is initially at rest when a
player uses a cue to increase its speed to 4.2 m/s at constant ac-
celeration. The acceleration takes place over a 2.0 m distance, at
the end of which the cue loses contact with the disk. Then the
disk slides an additional 12 m before stopping. Assume that the
shuffleboard court is level and that the force of friction on the
disk is constant.What is the increase in the thermal energy of the
diskcourt system (a) for that additional 12 m and (b) for the
entire 14 m distance? (c) How much work is done on the disk by
the cue?
123 A river descends 15 m through rapids. The speed of the wa-
ter is 3.2 m/s upon entering the rapids and 13 m/s upon leaving.
What percentage of the gravitational potential energy of the
water–Earth system is transferred to kinetic energy during the de-
scent? (Hint: Consider the descent of, say, 10 kg of water.)
SSM
131 Fasten one end of a vertical spring to a ceiling, attach a cab-
bage to the other end, and then slowly lower the cabbage until the
upward force on it from the spring balances the gravitational force
on it. Show that the loss of gravitational potential energy of the
cabbage–Earth system equals twice the gain in the spring’s poten-
tial energy.
132 The maximum force you can exert on an object with one of
your back teeth is about 750 N. Suppose that as you gradually bite
on a clump of licorice, the licorice resists compression by one of
your teeth by acting like a spring for which k2.5 105N/m. Find
(a) the distance the licorice is compressed by your tooth and
(b) the work the tooth does on the licorice during the compression.
(c) Plot the magnitude of your force versus the compression
distance. (d) If there is a potential energy associated with this com-
pression, plot it versus compression distance.
In the 1990s the pelvis of a particular Triceratops dinosaur was
found to have deep bite marks. The shape of the marks suggested
that they were made by a Tyrannosaurus rex dinosaur. To test the
idea, researchers made a replica of a T. rex tooth from bronze and
aluminum and then used a hydraulic press to gradually drive the
replica into cow bone to the depth seen in the Triceratops bone. A
graph of the force required versus depth of penetration is given in
Fig. 8-71 for one trial; the required force increased with depth be-
cause, as the nearly conical tooth penetrated the bone, more of the
tooth came in contact with the bone. (e) How much work was done
by the hydraulic press—and thus presumably by the T. rex—in
such a penetration? (f) Is there a potential energy associated with
this penetration? (The large biting force and energy expenditure
attributed to the T. rex by this research suggest that the animal was
a predator and not a scavenger.)
133 Conservative force F(x)
acts on a particle that moves
along an xaxis. Figure 8-72
shows how the potential energy
U(x) associated with force F(x)
varies with the position of the
particle, (a) Plot F(x) for the
range 0 x6 m. (b) The me-
chanical energy Eof the system
is 4.0 J. Plot the kinetic energy
K(x) of the particle directly on
Fig. 8-72.
134 Figure 8-73ashows a mol-
ecule consisting of two atoms of
masses mand M(with mM)
and separation r. Figure 8-73b
shows the potential energy U(r)
of the molecule as a function of
r. Describe the motion of the
atoms (a) if the total mechanical
energy Eof the two-atom sys-
tem is greater than zero (as is
E1), and (b) if Eis less than zero
(as is E2). For E111019 J
and r0.3 nm, find (c) the po-
tential energy of the system, (d)
the total kinetic energy of the
atoms, and (e) the force (magni-
tude and direction) acting on
each atom. For what values of r
is the force (f) repulsive, (g) at-
tractive, and (h) zero?
135 Repeat Problem 83, but now with the block accelerated up a
frictionless plane inclined at 5.0to the horizontal.
136 A spring with spring constant k620 N/m is placed in a ver-
tical orientation with its lower end supported by a horizontal sur-
face. The upper end is depressed 25 cm, and a block with a weight
of 50 N is placed (unattached) on the depressed spring.The system
is then released from rest. Assume that the gravitational potential
energy Ugof the block is zero at the release point (y0) and cal-
culate the kinetic energy Kof the block for yequal to (a) 0,
(b) 0.050 m, (c) 0.10 m, (d) 0.15 m, and (e) 0.20 m.Also, (f) how far
above its point of release does the block rise?
213
PROBLEMS
0
0
2000
4000
6000
8000
1 2 3 4 5 6
Penetration depth (mm)
7 8 9 10 11 12
Force (N)
Figure 8-71 Problem 132.
U(x) ( J)
3
x
(
m
)
4 5 6 10
4
3
2
1
2
Figure 8-72 Problem 133.
U(r), E (10
1
9J)
3
2
1
0
–1
–2
–3
0.2
r (nm)
(
b
)
E1
E2
r
m
M
0.3 0.4 0.10
(a)
Figure 8-73 Problem 134.
214
What Is Physics?
Every mechanical engineer who is hired as a courtroom expert witness to recon-
struct a traffic accident uses physics. Every dance trainer who coaches a ballerina
on how to leap uses physics. Indeed, analyzing complicated motion of any sort re-
quires simplification via an understanding of physics. In this chapter we discuss
how the complicated motion of a system of objects, such as a car or a ballerina,
can be simplified if we determine a special point of the systemthe center of
mass of that system.
Here is a quick example. If you toss a ball into the air without much spin on the
ball (Fig. 9-1a), its motion is simpleit follows a parabolic path, as we discussed in
Chapter 4, and the ball can be treated as a particle. If, instead, you flip a baseball bat
into the air (Fig.9-1b), its motion is more complicated. Because every part of the bat
moves differently, along paths of many different shapes, you cannot represent the
bat as a particle. Instead, it is a system of particles each of which follows its own path
through the air. However, the bat has one special pointthe center of massthat
does move in a simple parabolic path. The other parts of the bat move around the
center of mass. (To locate the center of mass, balance the bat on an outstretched fin-
ger;the point is above your finger, on the bat’s central axis.)
You cannot make a career of flipping baseball bats into the air, but you can
make a career of advising long-jumpers or dancers on how to leap properly into
the air while either moving their arms and legs or rotating their torso. Your
starting point would be to determine the person’s center of mass because of its
simple motion.
CHAPTER 9
Center of Mass and Linear Momentum
9-1 CENTER OF MASS
After reading this module, you should be able to . . .
9.01 Given the positions of several particles along an axis or
a plane, determine the location of their center of mass.
9.02 Locate the center of mass of an extended, symmetric
object by using the symmetry.
9.03 For a two-dimensional or three-dimensional extended ob-
ject with a uniform distribution of mass, determine the center
of mass by (a) mentally dividing the object into simple geomet-
ric figures, each of which can be replaced by a particle at its
center and (b) finding the center of mass of those particles.
The center of mass of a system of nparticles is defined to be the point whose coordinates are given by
or
where Mis the total mass of the system.
r
:
com 1
M
n
i1
mir
:
i,
xcom 1
M
n
i1
mixi, ycom 1
M
n
i1
miyi, zcom 1
M
n
i1
mizi,
Key Idea
Learning Objectives
215
9-1 CENTER OF MASS
The Center of Mass
We define the center of mass (com) of a system of particles (such as a person) in
order to predict the possible motion of the system.
Here we discuss how to determine where the center of mass of a system of parti-
cles is located.We start with a system of only a few particles, and then we consider
a system of a great many particles (a solid body, such as a baseball bat). Later in
the chapter, we discuss how the center of mass of a system moves when external
forces act on the system.
Systems of Particles
Two Particles. Figure 9-2ashows two particles of masses m1and m2separated by dis-
tance d.We have arbitrarily chosen the origin of an xaxis to coincide with the particle
of mass m1.We define the position of the center of mass (com) of this two-particle sys-
tem to be
(9-1)
Suppose, as an example, that m20.Then there is only one particle, of mass m1,
and the center of mass must lie at the position of that particle;Eq. 9-1 dutifully reduces
to xcom 0. If m10,there is again only one particle (of mass m2), and we have, as we
expect, xcom d. If m1m2, the center of mass should be halfway between the two
particles; Eq. 9-1 reduces to again as we expect. Finally, Eq. 9-1 tells us that
if neither m1nor m2is zero, xcom can have only values that lie between zero and d; that
is, the center of mass must lie somewhere between the two particles.
We are not required to place the origin of the coordinate system on one of
the particles. Figure 9-2bshows a more generalized situation, in which the coordi-
nate system has been shifted leftward. The position of the center of mass is now
defined
as (9-2)
Note that if we put x10, then x2becomes dand Eq. 9-2 reduces to Eq. 9-1, as
it must. Note also that in spite of the shift of the coordinate system, the center
xcom m1x1m2x2
m1m2
.
xcom 1
2d,
xcom m2
m1m2
d.
Figure 9-1 (a) A ball tossed into the air
follows a parabolic path. (b) The center
of mass (black dot) of a baseball bat
flipped into the air follows a parabolic
path, but all other points of the bat
follow more complicated curved paths.
(a)
(
b
)
Richard Megna/Fundamental Photographs
The center of mass of a system of particles is the point that moves as though
(1) all of the system’s mass were concentrated there and (2) all external forces
were applied there.
Figure 9-2 (a) Two particles of masses m1and m2are separated by distance d. The dot
labeled com shows the position of the center of mass, calculated from Eq. 9-1. (b) The
same as (a) except that the origin is located farther from the particles. The position of
the center of mass is calculated from Eq. 9-2. The location of the center of mass with
respect to the particles is the same in both cases.
x
y
xcom
x1d
com
m1m2
x2
(b)
x
y
xcom
d
com
m1m2
(a)
This is the center of mass
of the two-particle system.
Shifting the axis
does not change
the relative position
of the com.
of mass is still the same distance from each particle. The com is a property of the
physical particles, not the coordinate system we happen to use.
We can rewrite Eq. 9-2 as
(9-3)
in which Mis the total mass of the system. (Here, Mm1m2.)
Many Particles. We can extend this equation to a more general situation in
which nparticles are strung out along the xaxis. Then the total mass is Mm1
m2mn, and the location of the center of mass is
(9-4)
The subscript iis an index that takes on all integer values from 1 to n.
Three Dimensions. If the particles are distributed in three dimensions, the cen-
ter of mass must be identified by three coordinates. By extension of Eq. 9-4, they are
(9-5)
We can also define the center of mass with the language of vectors. First
recall that the position of a particle at coordinates xi,yi, and ziis given by a posi-
tion vector (it points from the origin to the particle):
(9-6)
Here the index identifies the particle, and i
ˆ,ˆ
j, and ˆ
k are unit vectors pointing,
respectively, in the positive direction of the x,y, and zaxes. Similarly, the position
of the center of mass of a system of particles is given by a position vector:
(9-7)
If you are a fan of concise notation, the three scalar equations of Eq. 9-5 can now
be replaced by a single vector equation,
(9-8)
where again Mis the total mass of the system. You can check that this equation
is correct by substituting Eqs. 9-6 and 9-7 into it, and then separating out the x,
y, and zcomponents.The scalar relations of Eq. 9-5 result.
Solid Bodies
An ordinary object, such as a baseball bat, contains so many particles (atoms)
that we can best treat it as a continuous distribution of matter. The “particles”
then become differential mass elements dm, the sums of Eq. 9-5 become inte-
grals, and the coordinates of the center of mass are defined as
(9-9)
where Mis now the mass of the object.The integrals effectively allow us to use Eq.
9-5 for a huge number of particles, an effort that otherwise would take many years.
Evaluating these integrals for most common objects (such as a television set or
a moose) would be difficult, so here we consider only uniform objects. Such objects
have uniform density, or mass per unit volume; that is, the density r(Greek letter
xcom 1
Mxdm, ycom 1
Mydm, zcom 1
Mzdm,
rcom
:1
M
n
i1
miri
:,
rcom
:xcomˆ
iycomˆ
jzcom ˆ
k.
ri
:xii
ˆyij
ˆzik
ˆ.
xcom 1
M
n
i1
mixi, ycom 1
M
n
i1
miyi, zcom 1
M
n
i1
mizi.
1
M
n
i1
mixi.
xcom m1x1m2x2m3x3 mnxn
M

xcom m1x1m2x2
M,
216 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
rho) is the same for any given element of an object as for the whole object. From
Eq. 1-8,we can write
(9-10)
where dV is the volume occupied by a mass element dm, and Vis the total vol-
ume of the object. Substituting dm (M/V)dV from Eq. 9-10 into Eq. 9-9 gives
(9-11)
Symmetry as a Shortcut. You can bypass one or more of these integrals if
an object has a point, a line, or a plane of symmetry. The center of mass of such
an object then lies at that point, on that line, or in that plane. For example, the
center of mass of a uniform sphere (which has a point of symmetry) is at the
center of the sphere (which is the point of symmetry). The center of mass of a
uniform cone (whose axis is a line of symmetry) lies on the axis of the cone. The
center of mass of a banana (which has a plane of symmetry that splits it into two
equal parts) lies somewhere in the plane of symmetry.
The center of mass of an object need not lie within the object. There is no
dough at the com of a doughnut, and no iron at the com of a horseshoe.
xcom 1
VxdV, ycom 1
VydV, zcom 1
VzdV.
rdm
dV M
V,
217
9-1 CENTER OF MASS
sides (Fig. 9-3). The three particles then have the following
coordinates:
Particle Mass (kg) x(cm) y(cm)
1 1.2 0 0
2 2.5 140 0
3 3.4 70 120
The total mass Mof the system is 7.1 kg.
From Eq. 9-5, the coordinates of the center of mass are
(Answer)
and
(Answer)
In Fig. 9-3, the center of mass is located by the position vec-
tor , which has components xcom and ycom. If we had
chosen some other orientation of the coordinate system,
these coordinates would be different but the location of the
com relative to the particles would be the same.
r
:
com
58 cm.
(1.2 kg)(0) (2.5 kg)(0) (3.4 kg)(120 cm)
7.1 kg
ycom 1
M
3
i1
miyim1y1m2y2m3y3
M
83 cm
(1.2 kg)(0) (2.5 kg)(140 cm) (3.4 kg)(70 cm)
7.1 kg
xcom 1
M
3
i1
mixim1x1m2x2m3x3
M
Sample Problem 9.01 com of three particles
Three particles of masses m11.2 kg, m22.5 kg, and
m33.4 kg form an equilateral triangle of edge length
a140 cm.Where is the center of mass of this system?
KEY IDEA
We are dealing with particles instead of an extended solid
body, so we can use Eq. 9-5 to locate their center of mass.
The particles are in the plane of the equilateral triangle, so
we need only the first two equations.
Calculations: We can simplify the calculations by choosing
the xand yaxes so that one of the particles is located at the
origin and the xaxis coincides with one of the triangle’s
Figure 9-3 Three particles form an equilateral triangle of edge
length a. The center of mass is located by the position vector .r
:
com
y
x
0
50
1
00
1
50
50
100
150
y
com
xcom
m1
m2
m3
rcom
a a
0
This is the position
vector rcom for the
com (it points from
the origin to the com).
Additional examples, video, and practice available at WileyPLUS
218 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Center Location
Plate of Mass of com Mass
PcomPxP?mP
ScomSxSRm
S
CcomCxC0mCmSmP
Assume that mass mSof disk Sis concentrated in a parti-
cle at xSR, and mass mPis concentrated in a particle
at xP(Fig. 9-4d). Next we use Eq. 9-2 to find the center of
mass xSPof the two-particle system:
(9-12)
Next note that the combination of disk Sand plate Pis
composite plate C. Thus, the position xSPof comSPmust
coincide with the position xCof comC, which is at the origin; so
xSPxC0.Substituting this into Eq. 9-12,we get
(9-13)
We can relate these masses to the face areas of Sand Pby
noting that
mass density volume
density thickness area.
Then
Because the plate is uniform, the densities and thicknesses
are equal; we are left with
Substituting this and xSRinto Eq. 9-13, we have
(Answer)xP1
3R.
pR2
p(2R)2pR21
3.
mS
mP
areaS
areaP
areaS
areaCareaS
mS
mP
densityS
densityP
thicknessS
thicknessP
areaS
areaP
.
xPxS
mS
mP
.
xSPmSxSmPxP
mSmP
.
Sample Problem 9.02 com of plate with missing piece
This sample problem has lots of words to read, but they will
allow you to calculate a com using easy algebra instead of
challenging integral calculus. Figure 9-4ashows a uniform
metal plate Pof radius 2Rfrom which a disk of radius Rhas
been stamped out (removed) in an assembly line.The disk is
shown in Fig. 9-4b. Using the xy coordinate system shown,
locate the center of mass comPof the remaining plate.
KEY IDEAS
(1) Let us roughly locate the center of plate Pby using sym-
metry. We note that the plate is symmetric about the xaxis
(we get the portion below that axis by rotating the upper
portion about the axis). Thus, comPmust be on the xaxis.
The plate (with the disk removed) is not symmetric about
the yaxis. However, because there is somewhat more mass
on the right of the yaxis, comPmust be somewhat to the
right of that axis. Thus, the location of comPshould be
roughly as indicated in Fig.9-4a.
(2) Plate Pis an extended solid body, so in principle we
can use Eqs. 9-11 to find the actual coordinates of the center
of mass of plate P. Here we want the xy coordinates of the
center of mass because the plate is thin and uniform. If it
had any appreciable thickness, we would just say that the
center of mass is midway across the thickness. Still, using
Eqs. 9-11 would be challenging because we would need a
function for the shape of the plate with its hole, and then we
would need to integrate the function in two dimensions.
(3) Here is a much easier way: In working with centers
of mass, we can assume that the mass of a uniform object (as
we have here) is concentrated in a particle at the object’s
center of mass.Thus we can treat the object as a particle and
avoid any two-dimensional integration.
Calculations: First, put the stamped-out disk (call it disk S)
back into place (Fig. 9-4c) to form the original composite
plate (call it plate C). Because of its circular symmetry, the
center of mass comSfor disk Sis at the center of S, at x
R(as shown). Similarly, the center of mass comCfor com-
posite plate Cis at the center of C, at the origin (as shown).
We then have the following:
Additional examples, video, and practice available at WileyPLUS
Checkpoint 1
The figure shows a uniform square plate from which four identical
squares at the corners will be removed. (a) Where is the center of mass of
the plate originally? Where is it after the removal of (b) square 1; (c)
squares 1 and 2; (d) squares 1 and 3; (e) squares 1,2, and 3; (f) all four
squares? Answer in terms of quadrants, axes,or points (without calcula-
tion, of course).
y
x
1 2
4 3
219
9-1 CENTER OF MASS
A
The com of the composite
plate is the same as the
com of the two pieces.
Plate P
2R
R
y
x
y
y
x
comP
comC
comS
Disk S
Composite plate
C = S + P
(a)
(b)
(c)
(d)x
comP
comC
comS
Disk particle Plate particle
Assume the plate's
mass is concentrated
as a particle at the
plate's center of mass.
Here too, assume the
mass is concentrated
as a particle at the
center of mass.
Here too.
Here are those
three particles.
Figure 9-4 (a) Plate Pis a metal plate of radius 2R, with a circular hole of radius R.The center of mass of Pis at point comP.(b) Disk S.
(c) Disk Shas been put back into place to form a composite plate C. The center of mass comSof disk Sand the center of mass comC
of plate Care shown. (d) The center of mass comSPof the combination of Sand Pcoincides with comC, which is at x0.
Newton’s Second Law for a System of Particles
Now that we know how to locate the center of mass of a system of particles, we
discuss how external forces can move a center of mass. Let us start with a simple
system of two billiard balls.
If you roll a cue ball at a second billiard ball that is at rest, you expect that the
two-ball system will continue to have some forward motion after impact. You
would be surprised, for example, if both balls came back toward you or if both
moved to the right or to the left. You already have an intuitive sense that some-
thing continues to move forward.
What continues to move forward, its steady motion completely unaf-
fected by the collision, is the center of mass of the two-ball system. If you fo-
cus on this pointwhich is always halfway between these bodies because
they have identical massesyou can easily convince yourself by trial at a bil-
liard table that this is so. No matter whether the collision is glancing, head-on,
or somewhere in between, the center of mass continues to move forward, as if
the collision had never occurred. Let us look into this center-of-mass motion
in more detail.
Motion of a System’s com. To do so, we replace the pair of billiard balls with
a system of nparticles of (possibly) different masses. We are interested not in the
individual motions of these particles but only in the motion of the center of mass
of the system. Although the center of mass is just a point, it moves like a particle
whose mass is equal to the total mass of the system; we can assign a position, a ve-
locity, and an acceleration to it. We state (and shall prove next) that the vector
equation that governs the motion of the center of mass of such a system of parti-
cles is
(system of particles). (9-14)
This equation is Newton’s second law for the motion of the center of mass of
a system of particles. Note that its form is the same as the form of the equation
F
:
net Ma
:
com
220 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES
After reading this module, you should be able to . . .
9.04 Apply Newton’s second law to a system of particles by re-
lating the net force (of the forces acting on the particles) to
the acceleration of the system’s center of mass.
9.05 Apply the constant-acceleration equations to the motion
of the individual particles in a system and to the motion of
the system’s center of mass.
9.06 Given the mass and velocity of the particles in a system,
calculate the velocity of the system’s center of mass.
9.07 Given the mass and acceleration of the particles in a
system, calculate the acceleration of the system’s center
of mass.
9.08 Given the position of a system’s center of mass as a func-
tion of time, determine the velocity of the center of mass.
9.09 Given the velocity of a system’s center of mass as a
function of time, determine the acceleration of the center
of mass.
9.10 Calculate the change in the velocity of a com by integrat-
ing the com’s acceleration function with respect to time.
9.11 Calculate a com’s displacement by integrating the
com’s velocity function with respect to time.
9.12 When the particles in a two-particle system move with-
out the system’s com moving, relate the displacements of
the particles and the velocities of the particles.
The motion of the center of mass of any system of particles
is governed by Newton’s second law for a system of parti-
cles, which is
.F
:
net Ma
:
com
Here is the net force of all the external forces acting on
the system, Mis the total mass of the system, and is the
acceleration of the system’s center of mass.
a
:com
F
:
net
Learning Objectives
Key Idea
for the motion of a single particle. However, the three quantities that
appear in Eq. 9-14 must be evaluated with some care:
1. is the net force of all external forces that act on the system. Forces on one
part of the system from another part of the system (internal forces) are not in-
cluded in Eq. 9-14.
2. Mis the total mass of the system. We assume that no mass enters or leaves the
system as it moves, so that Mremains constant. The system is said to be closed.
3. is the acceleration of the center of mass of the system. Equation 9-14 gives
no information about the acceleration of any other point of the system.
Equation 9-14 is equivalent to three equations involving the components of
and along the three coordinate axes.These equations are
Fnet,xMacom,xFnet,yMacom,yFnet,zMacom,z. (9-15)
Billiard Balls. Now we can go back and examine the behavior of the billiard
balls. Once the cue ball has begun to roll, no net external force acts on the (two-
ball) system. Thus, because 0, Eq. 9-14 tells us that 0 also. Because
acceleration is the rate of change of velocity, we conclude that the velocity of the
center of mass of the system of two balls does not change.When the two balls col-
lide, the forces that come into play are internal forces, on one ball from the other.
Such forces do not contribute to the net force , which remains zero. Thus, the
center of mass of the system, which was moving forward before the collision,
must continue to move forward after the collision, with the same speed and in the
same direction.
Solid Body. Equation 9-14 applies not only to a system of particles but also
to a solid body, such as the bat of Fig. 9-1b. In that case, Min Eq. 9-14 is the mass
of the bat and is the gravitational force on the bat. Equation 9-14 then tells us
that In other words, the center of mass of the bat moves as if the bat
were a single particle of mass M, with force acting on it.
Exploding Bodies. Figure 9-5 shows another interesting case. Suppose that at
a fireworks display, a rocket is launched on a parabolic path. At a certain point, it
explodes into fragments. If the explosion had not occurred, the rocket would have
continued along the trajectory shown in the figure. The forces of the explosion are
internal to the system (at first the system is just the rocket, and later it is its frag-
ments); that is, they are forces on parts of the system from other parts. If we ignore
air drag, the net external force acting on the system is the gravitational force on
the system, regardless of whether the rocket explodes. Thus, from Eq. 9-14, the ac-
celeration of the center of mass of the fragments (while they are in flight) re-
mains equal to This means that the center of mass of the fragments follows the
same parabolic trajectory that the rocket would have followed had it not exploded.
Ballet Leap. When a ballet dancer leaps across the stage in a grand jeté, she
raises her arms and stretches her legs out horizontally as soon as her feet leave the
g
:.
a
:
com
F
:
net
F
:
g
a
:
com g
:.
F
:
net
F
:
net
a
:
com
F
:
net
a
:
com
F
:
net
a
:
com
F
:
net
(Fnet
:ma
:)
221
9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES
Figure 9-5 A fireworks rocket explodes in
flight. In the absence of air drag, the center
of mass of the fragments would continue to
follow the original parabolic path, until
fragments began to hit the ground.
The internal forces of the
explosion cannot change
the path of the com.
stage (Fig. 9-6). These actions shift her center of mass upward through her body.
Although the shifting center of mass faithfully follows a parabolic path across the
stage, its movement relative to the body decreases the height that is attained by her
head and torso, relative to that of a normal jump.The result is that the head and torso
follow a nearly horizontal path,giving an illusion that the dancer is floating.
Proof of Equation 9-14
Now let us prove this important equation. From Eq. 9-8 we have, for a system of n
particles,
(9-16)
in which Mis the system’s total mass and is the vector locating the position of
the system’s center of mass.
Differentiating Eq. 9-16 with respect to time gives
(9-17)
Here is the velocity of the ith particle, and is the
velocity of the center of mass.
Differentiating Eq. 9-17 with respect to time leads to
(9-18)
Here is the acceleration of the ith particle, and is
the acceleration of the center of mass. Although the center of mass is just a geo-
metrical point, it has a position, a velocity, and an acceleration, as if it were a particle.
From Newton’s second law, is equal to the resultant force that acts on
the ith particle.Thus, we can rewrite Eq. 9-18 as
(9-19)
Among the forces that contribute to the right side of Eq. 9-19 will be forces that
the particles of the system exert on each other (internal forces) and forces
exerted on the particles from outside the system (external forces). By Newton’s
third law, the internal forces form third-law force pairs and cancel out in the sum
that appears on the right side of Eq. 9-19. What remains is the vector sum of
all the external forces that act on the system. Equation 9-19 then reduces to
Eq. 9-14, the relation that we set out to prove.
Ma
:
com F1
:F2
:F3
: Fn
:.
Fi
:
miai
:
a
:
com (dv
:com /dt)a
:
i (dv
:
i/dt)
Ma
:
com m1a1
:m2a2
:m3a3
: mnan
:.
v
:com (dr
:com /dt)vi
: (dr
i
:/dt)
Mv
:com m1v1
:m2v2
:m3v3
:mnvn
:.
rcom
:
Mr
:
com m1r1
:m2r2
:m3r3
: mnrn
:,
222 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Path of head
Path of center of mass
Figure 9-6 A grand jeté. (Based on The Physics of Dance, by Kenneth Laws, Schirmer
Books, 1984.)
223
9-2 NEWTON’S SECOND LAW FOR A SYSTEM OF PARTICLES
Checkpoint 2
Two skaters on frictionless ice hold opposite ends of a pole of negligible mass.An axis
runs along it, with the origin at the center of mass of the two-skater system. One skater,
Fred, weighs twice as much as the other skater, Ethel.Where do the skaters meet if (a)
Fred pulls hand over hand along the pole so as to draw himself to Ethel, (b) Ethel pulls
hand over hand to draw herself to Fred, and (c) both skaters pull hand over hand?
Sample Problem 9.03 Motion of the com of three particles
If the particles in a system all move together, the com moves
with them—no trouble there. But what happens when they
move in different directions with different accelerations?
Here is an example.
The three particles in Fig. 9-7aare initially at rest. Each
experiences an external force due to bodies outside the
three-particle system. The directions are indicated, and the
magnitudes are F16.0 N, F212 N, and F314 N. What
is the acceleration of the center of mass of the system, and in
what direction does it move?
KEY IDEAS
The position of the center of mass is marked by a dot in the
figure. We can treat the center of mass as if it were a real
particle, with a mass equal to the system’s total mass M16 kg.
We can also treat the three external forces as if they act at the
center of mass (Fig.9-7b).
Calculations: We can now apply Newton’s second law
to the center of mass, writing
(9-20)
or
so (9-21)
Equation 9-20 tells us that the acceleration of the
center of mass is in the same direction as the net external force
on the system (Fig. 9-7b). Because the particles are ini-
tially at rest, the center of mass must also be at rest. As the
center of mass then begins to accelerate, it must move off in
the common direction of and
We can evaluate the right side of Eq. 9-21 directly on
a vector-capable calculator, or we can rewrite Eq. 9-21 in
component form, find the components of and then find
Along the xaxis, we have
6.0 N (12 N) cos 4514 N
16 kg 1.03 m/s2.
acom, xF1xF2xF3x
M
a
:
com.
a
:
com,
F
:
net.a
:
com
F
:
net
a
:
com
a
:
com F1
:F2
:F3
:
M.
F1
:F2
:F3
:Ma
:
com
F
:
net Ma
:com
(F
:
net ma
:)
Figure 9-7 (a) Three particles, initially at rest in the positions shown,
are acted on by the external forces shown.The center of mass (com)
of the system is marked. (b) The forces are now transferred to the
center of mass of the system, which behaves like a particle with a
mass Mequal to the total mass of the system.The net external force
and the acceleration of the center of mass are shown.
a
:
com
F
:
net
x
y
3
2
1
0
–1
–2
–3
–3 –2 –1 1 2 3 4 5
x
y
3
2
1
0
–3 –2 –1 1 2 3 4 5
45°
8.0 kg
com
4.0 kg
4.0 kg
com
θ
M = 16 kg
(b)
(a)
F1F2
F3
F3
F1
F2Fnet
acom
The com of the system
will move as if all the
mass were there and
the net force acted there.
Additional examples, video, and practice available at WileyPLUS
Along the yaxis, we have
From these components, we find that has the magnitude
(Answer)
and the angle (from the positive direction of the xaxis)
(Answer)
tan1acom, y
acom, x
27.
1.16 m/s21.2 m/s2
acom 2(acom, x)2(acom, y)2
a
:
com
0(12 N) sin 450
16 kg 0.530 m/s2.
acom, yF1yF2yF3y
M
Linear Momentum
Here we discuss only a single particle instead of a system of particles, in order to
define two important quantities. Then we shall extend those definitions to sys-
tems of many particles.
The first definition concerns a familiar wordmomentum that has several
meanings in everyday language but only a single precise meaning in physics and
engineering. The linear momentum of a particle is a vector quantity that is
defined as
(linear momentum of a particle), (9-22)
in which mis the mass of the particle and is its velocity. (The adjective linear is of-
ten dropped, but it serves to distinguish from angular momentum, which is intro-
duced in Chapter 11 and which is associated with rotation.) Since mis always a
positive scalar quantity, Eq. 9-22 tells us that and have the same direction. Fromv
:
p
:
p
:
v
:
p
:mv
:
p
:
224 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
9-3 LINEAR MOMENTUM
After reading this module, you should be able to . . .
9.13 Identify that momentum is a vector quantity and thus has
both magnitude and direction and also components.
9.14 Calculate the (linear) momentum of a particle as the
product of the particle’s mass and velocity.
9.15 Calculate the change in momentum (magnitude and di-
rection) when a particle changes its speed and direction of
travel.
9.16 Apply the relationship between a particle’s momentum
and the (net) force acting on the particle.
9.17 Calculate the momentum of a system of particles as the
product of the system’s total mass and its center-of-mass
velocity.
9.18 Apply the relationship between a system’s center-of-
mass momentum and the net force acting on the system.
For a single particle, we define a quantity called its linear
momentum as
,
which is a vector quantity that has the same direction as the
particle’s velocity. We can write Newton’s second law in
p
:mv
:
p
:terms of this momentum:
For a system of particles these relations become
and F
:
net dP
:
dt .P
:Mv
:
com
F
:
net dp
:
dt .
Learning Objectives
Key Ideas
The time rate of change of the momentum of a particle is equal to the net force
acting on the particle and is in the direction of that force.
In equation form this becomes
(9-23)
In words, Eq. 9-23 says that the net external force on a particle changes the
particle’s linear momentum Conversely, the linear momentum can be
changed only by a net external force. If there is no net external force, cannot
change.As we shall see in Module 9-5, this last fact can be an extremely power-
ful tool in solving problems.
p
:
p
:.
F
:
net
F
:
net dp
:
dt .
Eq.9-22, the SI unit for momentum is the kilogram-meter per second (kg m/s).
Force and Momentum. Newton expressed his second law of motion in terms
of momentum:
Manipulating Eq. 9-23 by substituting for from Eq. 9-22 gives, for constant
mass m,
Thus, the relations and are equivalent expressions of
Newton’s second law of motion for a particle.
F
:
net ma
:
F
:
net dp
:/dt
F
:
net dp
:
dt d
dt (mv
:)mdv
:
dt ma
:.
p
:
225
9-3 LINEAR MOMENTUM
Checkpoint 3
The figure gives the magnitude pof the linear mo-
mentum versus time tfor a particle moving along
an axis.A force directed along the axis acts on the
particle.(a) Rank the four regions indicated ac-
cording to the magnitude of the force,greatest
first.(b) In which region is the particle slowing?
The Linear Momentum of a System of Particles
Let’s extend the definition of linear momentum to a system of particles. Consider
a system of nparticles, each with its own mass, velocity, and linear momentum.
The particles may interact with each other, and external forces may act on them.
The system as a whole has a total linear momentum which is defined to be the
vector sum of the individual particles’ linear momenta.Thus,
(9-24)
If we compare this equation with Eq. 9-17, we see that
(linear momentum, system of particles), (9-25)
which is another way to define the linear momentum of a system of particles:
P
:Mv
:com
m1v
:1m2v
:2m3v
:3 mnv
:
n.
P
:p
:1p
:2p
:3 p
:
n
P
:
,
The linear momentum of a system of particles is equal to the product of the total
mass Mof the system and the velocity of the center of mass.
Force and Momentum. If we take the time derivative of Eq. 9-25 (the veloc-
ity can change but not the mass), we find
(9-26)
Comparing Eqs. 9-14 and 9-26 allows us to write Newton’s second law for a sys-
tem of particles in the equivalent form
(system of particles), (9-27)
where is the net external force acting on the system.This equation is the gen-
eralization of the single-particle equation to a system of many
particles. In words, the equation says that the net external force on a system
of particles changes the linear momentum of the system. Conversely, the linear
momentum can be changed only by a net external force. If there is no net exter-
nal force, cannot change. Again, this fact gives us an extremely powerful tool
for solving problems.
P
:
P
:F
:
net
F
:
net dp
:/dt
F
:
net
F
:
net dP
:
dt
dP
:
dt Mdv
:com
dt Ma
:com.
p
t
1
2
3
4
226 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
9-4 COLLISION AND IMPULSE
After reading this module, you should be able to . . .
9.19 Identify that impulse is a vector quantity and thus has both
magnitude and direction and also components.
9.20 Apply the relationship between impulse and momentum
change.
9.21 Apply the relationship between impulse, average force,
and the time interval taken by the impulse.
9.22 Apply the constant-acceleration equations to relate im-
pulse to average force.
9.23 Given force as a function of time, calculate the impulse (and
thus also the momentum change) by integrating the function.
9.24 Given a graph of force versus time, calculate the im-
pulse (and thus also the momentum change) by graphical
integration.
9.25 In a continuous series of collisions by projectiles, calcu-
late the average force on the target by relating it to the rate
at which mass collides and to the velocity change experi-
enced by each projectile.
Applying Newton’s second law in momentum form to a
particle-like body involved in a collision leads to the
impulse–linear momentum theorem:
,
where is the change in the body’s linear momen-p
:
fp
:
ip
:
p
:
fp
:
ip
:J
:
When a steady stream of bodies, each with mass mand
speed v, collides with a body whose position is fixed, the aver-
age force on the fixed body is
where n/tis the rate at which the bodies collide with the
fixed body, and vis the change in velocity of each colliding
body. This average force can also be written as
where m/tis the rate at which mass collides with the fixed
body. The change in velocity is vvif the bodies stop
upon impact and v2vif they bounce directly backward
with no change in their speed.
Favg m
tv,
Favg  n
tp n
tmv,
Learning Objectives
Key Ideas
tum, and is the impulse due to the force exerted on the
body by the other body in the collision:
If Favg is the average magnitude of during the collisionF
:
(t)
J
:tf
ti
F
:
(t)dt.
F
:
(t)J
:
and tis the duration of the collision, then for one-dimensional
motion
JFavg t.
Collision and Impulse
The momentum of any particle-like body cannot change unless a net
external force changes it. For example, we could push on the body to change its
momentum. More dramatically, we could arrange for the body to collide with a
baseball bat. In such a collision (or crash), the external force on the body is brief,
has large magnitude, and suddenly changes the body’s momentum. Collisions oc-
cur commonly in our world, but before we get to them, we need to consider a sim-
ple collision in which a moving particle-like body (a projectile) collides with some
other body (a target).
Single Collision
Let the projectile be a ball and the target be a bat.The collision is brief, and the ball
experiences a force that is great enough to slow, stop, or even reverse its motion.
Figure 9-8 depicts the collision at one instant.The ball experiences a force that
varies during the collision and changes the linear momentum of the ball. Thatp
:
F
:
(t)
p
:
The collision of a ball with a bat collapses
part of the ball.
Photo by Harold E. Edgerton. © The Harold and Esther Edgerton
Family Trust, courtesy of Palm Press, Inc.
change is related to the force by Newton’s second law written in the form
By rearranging this second-law expression, we see that, in time interval dt, the
change in the ball’s momentum is
(9-28)dp
:F
:
(t)dt.
F
:dp
:/dt.
We can find the net change in the ball’s momentum due to the collision if we inte-
grate both sides of Eq. 9-28 from a time tijust before the collision to a time tfjust
after the collision:
(9-29)
The left side of this equation gives us the change in momentum:
The right side, which is a measure of both the magnitude and the duration of the
collision force, is called the impulse of the collision:
(impulse defined). (9-30)
Thus, the change in an object’s momentum is equal to the impulse on the object:
(linear momentumimpulse theorem). (9-31)
This expression can also be written in the vector form
(9-32)
and in such component forms as
pxJx(9-33)
and (9-34)
Integrating the Force. If we have a function for we can evaluate (and
thus the change in momentum) by integrating the function. If we have a plot of
versus time t, we can evaluate by finding the area between the curve and the t
axis, such as in Fig. 9-9a. In many situations we do not know how the force varies
with time but we do know the average magnitude Favg of the force and the duration
t(tfti) of the collision. Then we can write the magnitude of the impulse as
JFavg t. (9-35)
The average force is plotted versus time as in Fig. 9-9b.The area under that curve
is equal to the area under the curve for the actual force F(t) in Fig. 9-9abecause
both areas are equal to impulse magnitude J.
Instead of the ball, we could have focused on the bat in Fig. 9-8. At any
instant, Newton’s third law tells us that the force on the bat has the same
magnitude but the opposite direction as the force on the ball. From Eq. 9-30, this
means that the impulse on the bat has the same magnitude but the opposite
direction as the impulse on the ball.
J
:F
:
J
:
F
:(t),
pfx pix tf
ti
Fxdt.
p
:
fp
:
iJ
:
p
:J
:
J
:tf
ti
F
:
(t)dt
J
:
p
:
fp
:
ip
:.
tf
ti
dp
:tf
ti
F
:
(t)dt.
227
9-4 COLLISION AND IMPULSE
Figure 9-8 Force acts on a ball as the
ball and a bat collide.
F
:(t)
x
Bat Ball
F(t)
Figure 9-9 (a) The curve shows the magni-
tude of the time-varying force F(t) that acts
on the ball in the collision of Fig. 9-8. The
area under the curve is equal to the magni-
tude of the impulse on the ball in the col-
lision. (b) The height of the rectangle repre-
sents the average force Favg acting on the
ball over the time interval t.The area within
the rectangle is equal to the area under the
curve in (a) and thus is also equal to the
magnitude of the impulse in the collision.J
:
J
:
ti
F
J
F(t)
tf
Δt
Δt
t
ti
F
Favg
tf
t
J
(a)
(b)
The impulse in the collision
is equal to the area under
the curve.
The average force gives
the same area under the
curve.
Checkpoint 4
A paratrooper whose chute fails to open lands in snow; he is hurt slightly. Had he
landed on bare ground, the stopping time would have been 10 times shorter and the
collision lethal. Does the presence of the snow increase,decrease, or leave unchanged
the values of (a) the paratrooper’s change in momentum,(b) the impulse stopping the
paratrooper,and (c) the force stopping the paratrooper?
Series of Collisions
Now let’s consider the force on a body when it undergoes a series of identical, re-
peated collisions. For example, as a prank, we might adjust one of those machines
that fire tennis balls to fire them at a rapid rate directly at a wall. Each collision
would produce a force on the wall, but that is not the force we are seeking. We
want the average force Favg on the wall during the bombardmentthat is, the av-
erage force during a large number of collisions.
In Fig. 9-10, a steady stream of projectile bodies, with identical mass mand
linear momenta moves along an xaxis and collides with a target body that ismv
:,
228 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Figure 9-10 A steady stream of projectiles,
with identical linear momenta, collides
with a target, which is fixed in place. The
average force Favg on the target is to the
right and has a magnitude that depends on
the rate at which the projectiles collide
with the target or, equivalently, the rate at
which mass collides with the target.
x
Target
v
Projectiles
fixed in place. Let nbe the number of projectiles that collide in a time interval t.
Because the motion is along only the xaxis, we can use the components of the
momenta along that axis. Thus, each projectile has initial momentum mv and
undergoes a change pin linear momentum because of the collision. The total
change in linear momentum for nprojectiles during interval tis np.The
resulting impulse on the target during tis along the xaxis and has the same
magnitude of npbut is in the opposite direction. We can write this relation in
component form as
Jnp, (9-36)
where the minus sign indicates that Jand phave opposite directions.
Average Force. By rearranging Eq. 9-35 and substituting Eq. 9-36, we find
the average force Favg acting on the target during the collisions:
(9-37)
This equation gives us Favg in terms of n/t, the rate at which the projectiles
collide with the target, and v, the change in the velocity of those projectiles.
Velocity Change. If the projectiles stop upon impact, then in Eq. 9-37 we can
substitute, for v,
vvfvi0vv, (9-38)
where vi(v) and vf(0) are the velocities before and after the collision,
respectively. If, instead, the projectiles bounce (rebound) directly backward from
the target with no change in speed, then vfvand we can substitute
vvfvivv2v. (9-39)
In time interval t, an amount of mass mnm collides with the target.
With this result, we can rewrite Eq. 9-37 as
(9-40)
This equation gives the average force Favg in terms of m/t, the rate at which
mass collides with the target. Here again we can substitute for vfrom Eq. 9-38
or 9-39 depending on what the projectiles do.
Favg m
tv.
Favg J
t n
tp n
tmv.
J
:
Checkpoint 5
The figure shows an overhead view of a ball bouncing from a vertical wall without any
change in its speed. Consider the change in the ball’s linear momentum.(a) Is px
positive, negative, or zero? (b) Is pypositive,negative, or zero? (c) What is the direc-
tion of ?p
:
p
:
θ
θ
y
x
229
9-4 COLLISION AND IMPULSE
Impulse: The impulse is then
(Answer)
which means the impulse magnitude is
The angle of is given by
(Answer)
which a calculator evaluates as 75.4. Recall that the physi-
cally correct result of an inverse tangent might be the
displayed answer plus 180.We can tell which is correct here
by drawing the components of (Fig. 9-11c). We find that u
is actually 75.4180255.4, which we can write as
u105. (Answer)
(b) The collision lasts for 14 ms. What is the magnitude of
the average force on the driver during the collision?
KEY IDEA
From Eq. 9-35 (JFavg t), the magnitude Favg of the aver-
age force is the ratio of the impulse magnitude Jto the dura-
tion tof the collision.
Calculations: We have
. (Answer)
Using Fma with m80 kg, you can show that the magni-
tude of the driver’s average acceleration during the collision
is about 3.22 103m/s2329g, which is fatal.
Surviving: Mechanical engineers attempt to reduce the
chances of a fatality by designing and building racetrack
walls with more “give, so that a collision lasts longer. For
example, if the collision here lasted 10 times longer and the
other data remained the same, the magnitudes of the aver-
age force and average acceleration would be 10 times less
and probably survivable.
2.583 105 N 2.6 105 N
Favg J
t3616 kgm/s
0.014 s
J
:
utan1Jy
Jx
,
J
:
J2Jx
2Jy
23616 kgm/s 3600 kgm/s.
J
:(910i
ˆ3500j
ˆ) kgm/s,
Sample Problem 9.04 Two-dimensional impulse, race car–wall collision
Figure 9-11ais an overhead view of
the path taken by a race car driver as his car collides with the
racetrack wall. Just before the collision, he is traveling at
speed vi70 m/s along a straight line at 30from the wall.
Just after the collision, he is traveling at speed vf50 m/s
along a straight line at 10from the wall. His mass mis 80 kg.
(a) What is the impulse on the driver due to the collision?
KEY IDEAS
We can treat the driver as a particle-like body and thus apply
the physics of this module. However, we cannot calculate
directly from Eq. 9-30 because we do not know anything about
the force on the driver during the collision. That is, we do
not have a function of or a plot for it and thus cannot
integrate to find . However,we can find from the change in
the driver’s linear momentum via Eq. 9-32 .
Calculations: Figure 9-11bshows the driver’s momentum p
:
i
(J
:p
:
fp
:
i)p
:
J
:
J
:F
:(t)
F
:(t)
J
:
J
:
Race carwall collision.
Wall
x
y
30°
10°
30°
Path
(a)
x
y
10°
(b)
pi
pf–105°
x
y
(c)
Jy
Jx
J
The impulse on the car
is equal to the change
in the momentum.
The collision
changes the
momentum.
Figure 9-11 (a) Overhead
view of the path taken by a
race car and its driver as the
car slams into the racetrack
wall. (b) The initial momen-
tum and final momentum
of the driver. (c) The
impulse on the driver
during the collision.
J
:
p
:
f
p
:
i
Additional examples, video, and practice available at WileyPLUS
before the collision (at angle 30from the positive xdirection)
and his momentum after the collision (at angle 10). From
Eqs. 9-32 and 9-22 ,we can write
(9-41)
We could evaluate the right side of this equation directly on
a vector-capable calculator because we know mis 80 kg,
is 50 m/s at 10, and is 70 m/s at 30. Instead, here we
evaluate Eq. 9-41 in component form.
x component: Along the xaxis we have
Jxm(vfx vix)
(80 kg)[(50 m/s) cos(10)(70 m/s) cos 30]
910 kgm/s.
y component: Along the yaxis,
Jym(vfy viy)
(80 kg)[(50 m/s) sin(10)(70 m/s) sin 30]
3495 kgm/s 3500 kgm/s.
v
:
i
v
:
f
J
:p
:
fp
:
imv
:
fmvi
:m(v
:
fv
:
i).
mv
:)( p
:
p
:
f
Conservation of Linear Momentum
Suppose that the net external force (and thus the net impulse ) acting on a
system of particles is zero (the system is isolated) and that no particles leave or
enter the system (the system is closed). Putting in Eq. 9-27 then yields
, which means that
(closed, isolated system). (9-42)
In words,
P
:constant
dP
:/dt 0
F
:
net 0
J
:
F
:
net
230 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
9-5 CONSERVATION OF LINEAR MOMENTUM
After reading this module, you should be able to . . .
9.26 For an isolated system of particles, apply the conservation
of linear momenta to relate the initial momenta of the particles
to their momenta at a later instant.
9.27 Identify that the conservation of linear momentum can be
done along an individual axis by using components along
that axis, provided that there is no net external force com-
ponent along that axis.
If a system is closed and isolated so that no net external
force acts on it, then the linear momentum must be constant
even if there are internal changes:
(closed, isolated system).P
:constant
P
:
This conservation of linear momentum can also be written
in terms of the system’s initial momentum and its momentum
at some later instant:
(closed, isolated system),P
:
iP
:
f
Learning Objectives
Key Ideas
If no net external force acts on a system of particles, the total linear momentum
of the system cannot change.
P
:
This result is called the law of conservation of linear momentum and is an extremely
powerful tool in solving problems. In the homework we usually write the law as
(closed, isolated system). (9-43)
In words, this equation says that, for a closed, isolated system,
.
Caution: Momentum should not be confused with energy. In the sample problems
of this module, momentum is conserved but energy is definitely not.
Equations 9-42 and 9-43 are vector equations and, as such, each is equivalent
to three equations corresponding to the conservation of linear momentum in
three mutually perpendicular directions as in, say, an xyz coordinate system.
Depending on the forces acting on a system, linear momentum might be
conserved in one or two directions but not in all directions. However,
total linear momentum
at some initial time ti
total linear momentum
at some later time tf
P
:
iP
:
f
If the component of the net external force on a closed system is zero along an axis, then
the component of the linear momentum of the system along that axis cannot change.
In a homework problem, how can you know if linear momentum can be con-
served along,say, an xaxis? Check the force components along that axis. If the net of
any such components is zero, then the conservation applies.As an example, suppose
that you toss a grapefruit across a room. During its flight, the only external force act-
ing on the grapefruit (which we take as the system) is the gravitational force ,
which is directed vertically downward. Thus, the vertical component of the linear
F
:
g
momentum of the grapefruit changes, but since no horizontal external force acts on
the grapefruit,the horizontal component of the linear momentum cannot change.
Note that we focus on the external forces acting on a closed system.
Although internal forces can change the linear momentum of portions of the sys-
tem, they cannot change the total linear momentum of the entire system. For ex-
ample, there are plenty of forces acting between the organs of your body, but they
do not propel you across the room (thankfully).
The sample problems in this module involve explosions that are either one-
dimensional (meaning that the motions before and after the explosion are along
a single axis) or two-dimensional (meaning that they are in a plane containing
two axes). In the following modules we consider collisions.
231
9-5 CONSERVATION OF LINEAR MOMENTUM
We can relate the vMS to the known velocities with
.
In symbols, this gives us
vHS vrel vMS (9-47)
or vMS vHS vrel.
Substituting this expression for vMS into Eq. 9-46, and then
substituting Eqs. 9-45 and 9-46 into Eq. 9-44,we find
Mvi0.20M(vHS vrel)0.80MvHS,
which gives us
vHS vi0.20vrel,
or vHS 2100 km/h (0.20)(500 km/h)
2200 km/h. (Answer)
velocity of
hauler relative
to Sun
velocity of
hauler relative
to module
velocity of
module relative
to Sun
Sample Problem 9.05 One-dimensional explosion, relative velocity, space hauler
One-dimensional explosion: Figure 9-12ashows a space hauler
and cargo module, of total mass M, traveling along an xaxis in
deep space. They have an initial velocity of magnitude 2100
km/h relative to the Sun. With a small explosion, the hauler
ejects the cargo module, of mass 0.20M(Fig. 9-12b).The hauler
then travels 500 km/h faster than the module along the xaxis;
that is, the relative speed vrel between the hauler and the mod-
ule is 500 km/h.What then is the velocity of the hauler rela-
tive to the Sun?
KEY IDEA
Because the haulermodule system is closed and isolated,
its total linear momentum is conserved; that is,
, (9-44)
where the subscripts iand frefer to values before and after
the ejection, respectively. (We need to be careful here:
Although the momentum of the system does not change, the
momenta of the hauler and module certainly do.)
Calculations: Because the motion is along a single axis, we can
write momenta and velocities in terms of their xcomponents,
using a sign to indicate direction. Before the ejection, we have
PiMvi. (9-45)
Let vMS be the velocity of the ejected module relative to the
Sun.The total linear momentum of the system after the ejec-
tion is then
Pf(0.20M)vMS (0.80M)vHS, (9-46)
where the first term on the right is the linear momentum of
the module and the second term is that of the hauler.
P
:
iP
:
f
v
:
HS
v
:
i
Figure 9-12 (a) A space hauler, with a cargo module, moving at initial
velocity (b) The hauler has ejected the cargo module. Now the
velocities relative to the Sun are for the module and for the
hauler.
v
:
HS
v
:
MS
v
:
i.
(
a
)(
b
)
Cargo module
Hauler
0.20M
vMS vHS
vi
0.80M
xx
The explosive separation can change the momentum
of the parts but not the momentum of the system.
Additional examples, video, and practice available at WileyPLUS
Checkpoint 6
An initially stationary device lying on a frictionless floor explodes into two pieces,which
then slide across the floor,one of them in the positive xdirection. (a) What is the sum of
the momenta of the two pieces after the explosion? (b) Can the second piece move at an
angle to the xaxis? (c) What is the direction of the momentum of the second piece?
232 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Calculations: Linear momentum is also conserved along
the xaxis because there is no net external force acting on
the coconut and pieces along that axis.Thus we have
Pix Pfx, (9-49)
where Pix 0 because the coconut is initially at rest. To
get Pfx, we find the xcomponents of the final momenta,
using the fact that piece Amust have a mass of 0.50M
(M0.20M0.30M):
pfA,x 0.50MvfA,
pfB,x 0.20MvfB,x 0.20MvfB cos 50,
pfC,x 0.30MvfC,x 0.30MvfC cos 80.
Equation 9-49 for the conservation of momentum along the
xaxis can now be written as
Pix Pfx pfA,x pfB,x pfC,x.
Then, with vfC 5.0 m/s and vfB 9.64 m/s, we have
00.50MvfA 0.20M(9.64 m/s) cos 50
0.30M(5.0 m/s) cos 80,
from which we find
vfA 3.0 m/s. (Answer)
Sample Problem 9.06 Two-dimensional explosion, momentum, coconut
Two-dimensional explosion: A firecracker placed inside a
coconut of mass M, initially at rest on a frictionless floor,
blows the coconut into three pieces that slide across the floor.
An overhead view is shown in Fig. 9-13a. Piece C, with mass
0.30M, has final speed vfC 5.0 m/s.
(a) What is the speed of piece B, with mass 0.20M?
KEY IDEA
First we need to see whether linear momentum is con-
served. We note that (1) the coconut and its pieces form a
closed system, (2) the explosion forces are internal to that
system, and (3) no net external force acts on the system.
Therefore, the linear momentum of the system is conserved.
(We need to be careful here: Although the momentum of
the system does not change, the momenta of the pieces cer-
tainly do.)
Calculations: To get started, we superimpose an xy coordinate
system as shown in Fig. 9-13b, with the negative direction of the
xaxis coinciding with the direction of The xaxis is at 80v
:
fA.
Additional examples, video, and practice available at WileyPLUS
Figure 9-13 Three pieces of an
exploded coconut move off in
three directions along a
frictionless floor. (a) An over-
head view of the event. (b) The
same with a two-dimensional
axis system imposed.
with the direction of and 50with the direction of .
Linear momentum is conserved separately along each
axis. Let’s use the yaxis and write
Piy Pfy, (9-48)
where subscript irefers to the initial value (before the ex-
plosion), and subscript yrefers to the ycomponent of
or .
The component Piy of the initial linear momentum is
zero, because the coconut is initially at rest. To get an ex-
pression for Pfy, we find the ycomponent of the final linear
momentum of each piece, using the y-component version of
Eq. 9-22 ( pymvy):
pfA,y 0,
pfB,y 0.20MvfB,y 0.20MvfB sin 50,
pfC,y 0.30MvfC,y 0.30MvfC sin 80.
(Note that pfA,y 0 because of our nice choice of axes.)
Equation 9-48 can now be written as
Piy Pfy pfA,y pfB,y pfC,y.
Then, with vfC 5.0 m/s, we have
000.20MvfB sin 50(0.30M)(5.0 m/s) sin 80,
from which we find
vfB 9.64 m/s 9.6 m/s. (Answer)
(b) What is the speed of piece A?
Pf
:Pi
:
v
:
fB
v
:
fC
A
B
C
vfB
vfC
vfA
100°
130°
(a)
B
C
vfB
vfC
vfA
80°
(b)
x
y
50°
A
The explosive separation
can change the momentum
of the parts but not the
momentum of the system.
Momentum and Kinetic Energy in Collisions
In Module 9-4, we considered the collision of two particle-like bodies but focused
on only one of the bodies at a time. For the next several modules we switch our
focus to the system itself, with the assumption that the system is closed and iso-
lated. In Module 9-5, we discussed a rule about such a system: The total linear
momentum of the system cannot change because there is no net external force
to change it.This is a very powerful rule because it can allow us to determine the
results of a collision without knowing the details of the collision (such as how
much damage is done).
We shall also be interested in the total kinetic energy of a system of two col-
liding bodies. If that total happens to be unchanged by the collision, then the
kinetic energy of the system is conserved (it is the same before and after the
collision). Such a collision is called an elastic collision. In everyday collisions of
common bodies, such as two cars or a ball and a bat, some energy is always trans-
ferred from kinetic energy to other forms of energy, such as thermal energy or
energy of sound. Thus, the kinetic energy of the system is not conserved. Such a
collision is called an inelastic collision.
However, in some situations, we can approximate a collision of common bod-
ies as elastic. Suppose that you drop a Superball onto a hard floor. If the collision
between the ball and floor (or Earth) were elastic, the ball would lose no kinetic
energy because of the collision and would rebound to its original height.
However, the actual rebound height is somewhat short, showing that at least
some kinetic energy is lost in the collision and thus that the collision is somewhat
inelastic. Still, we might choose to neglect that small loss of kinetic energy to ap-
proximate the collision as elastic.
The inelastic collision of two bodies always involves a loss in the kinetic
energy of the system. The greatest loss occurs if the bodies stick together, in
which case the collision is called a completely inelastic collision. The collision of a
baseball and a bat is inelastic. However, the collision of a wet putty ball and a bat
is completely inelastic because the putty sticks to the bat.
P
:
233
9-6 MOMENTUM AND KINETIC ENERGY IN COLLISIONS
9-6 MOMENTUM AND KINETIC ENERGY IN COLLISIONS
After reading this module, you should be able to . . .
9.28 Distinguish between elastic collisions, inelastic collisions,
and completely inelastic collisions.
9.29 Identify a one-dimensional collision as one where the ob-
jects move along a single axis, both before and after the
collision.
9.30 Apply the conservation of momentum for an isolated
one-dimensional collision to relate the initial momenta of
the objects to their momenta after the collision.
9.31 Identify that in an isolated system, the momentum and
velocity of the center of mass are not changed even if the
objects collide.
In an inelastic collision of two bodies, the kinetic energy of
the two-body system is not conserved. If the system is closed
and isolated, the total linear momentum of the system must
be conserved, which we can write in vector form as
,
where subscripts iand frefer to values just before and just
after the collision, respectively.
If the motion of the bodies is along a single axis, the collision
is one-dimensional and we can write the equation in terms of
p
:1ip
:2ip
:1fp
:2f
velocity components along that axis:
m1v1im2v2im1v1fm2v2f.
If the bodies stick together, the collision is a completely
inelastic collision and the bodies have the same final veloc-
ity V(because they are stuck together).
The center of mass of a closed, isolated system of two col-
liding bodies is not affected by a collision. In particular, the ve-
locity of the center of mass cannot be changed by the
collision.
v
:com
Learning Objectives
Key Ideas
Inelastic Collisions in One Dimension
One-Dimensional Inelastic Collision
Figure 9-14 shows two bodies just before and just after they have a one-
dimensional collision. The velocities before the collision (subscript i) and after
the collision (subscript f) are indicated.The two bodies form our system, which is
closed and isolated.We can write the law of conservation of linear momentum for
this two-body system as
,
which we can symbolize as
(conservation of linear momentum). (9-50)
Because the motion is one-dimensional, we can drop the overhead arrows for
vectors and use only components along the axis, indicating direction with a sign.
Thus, from pmv, we can rewrite Eq. 9-50 as
m1v1im2v2im1v1fm2v2f. (9-51)
If we know values for,say, the masses, the initial velocities, and one of the final ve-
locities, we can find the other final velocity with Eq. 9-51.
One-Dimensional Completely Inelastic Collision
Figure 9-15 shows two bodies before and after they have a completely inelastic
collision (meaning they stick together).The body with mass m2happens to be ini-
tially at rest (v2i0).We can refer to that body as the target and to the incoming
body as the projectile. After the collision, the stuck-together bodies move with
velocity V. For this situation, we can rewrite Eq. 9-51 as
m1v1i(m1m2)V(9-52)
or . (9-53)
If we know values for, say, the masses and the initial velocity v1iof the projectile,
we can find the final velocity Vwith Eq. 9-53. Note that Vmust be less than v1ibe-
cause the mass ratio m1/(m1m2) must be less than unity.
Velocity of the Center of Mass
In a closed, isolated system, the velocity of the center of mass of the system
cannot be changed by a collision because, with the system isolated, there is no net
external force to change it. To get an expression for , let us return to the v
:
com
v
:
com
Vm1
m1m2
v1i
p
:
1ip
:
2ip
:
1fp
:
2f
total momentum P
:
i
before the collision
total momentum P
:
f
after the collision
234 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Figure 9-14 Bodies 1 and 2 move along an
xaxis, before and after they have an
inelastic collision.
m1m2
Before
Body 1 Body 2
x
v1iv2i
m1m2
After
x
v1fv2f
Here is the generic setup
for an inelastic collision.
Figure 9-15 A completely inelastic collision between
two bodies. Before the collision, the body with mass
m2is at rest and the body with mass m1moves
directly toward it. After the collision, the stuck-
together bodies move with the same velocity .V
:
m1
Projectile
m2
Target
x
x
V
v1i
After
Before
m1 + m2
v2i = 0
In a completely inelastic
collision, the bodies
stick together.
two-body system and one-dimensional collision of Fig. 9-14. From Eq. 9-25
, we can relate to the total linear momentum of that two-body
system by writing
. (9-54)
The total linear momentum is conserved during the collision; so it is given by
either side of Eq. 9-50. Let us use the left side to write
. (9-55)
Substituting this expression for in Eq. 9-54 and solving for give us
. (9-56)
The right side of this equation is a constant, and has that same constant value
before and after the collision.
For example, Fig. 9-16 shows, in a series of freeze-frames, the motion of the
center of mass for the completely inelastic collision of Fig. 9-15. Body 2 is the tar-
get, and its initial linear momentum in Eq. 9-56 is Body 1 is
the projectile, and its initial linear momentum in Eq. 9-56 is Note
that as the series of freeze-frames progresses to and then beyond the collision,
the center of mass moves at a constant velocity to the right. After the
collision, the common final speed Vof the bodies is equal to because then
the center of mass travels with the stuck-together bodies.
v
:
com
p
:
1im1v
:
1i.
p
:
2im2v
:
2i0.
v
:
com
v
:
com P
:
m1m2
p
:
1ip
:
2i
m1m2
v
:
com
P
:
P
:p
:1ip
:2i
P
:
P
:Mv
:
com (m1m2)v
:
com
P
:
v
:
com
(P
:Mv
:
com)
235
9-6 MOMENTUM AND KINETIC ENERGY IN COLLISIONS
x
m1
v1iv2i= 0
m2
m1+m2
V=vcom
Collision!
vcom
The com of the two
bodies is between
them and moves at a
constant velocity.
Here is the
incoming projectile.
The com moves at the
same velocity even after
the bodies stick together.
Here is the
stationary target.
Figure 9-16 Some freeze-frames of the two-body system
in Fig. 9-15, which undergoes a completely inelastic col-
lision. The system’s center of mass is shown in each
freeze-frame. The velocity of the center of mass is
unaffected by the collision. Because the bodies stick
together after the collision, their common velocity
must be equal to .v
:
com
V
:
v
:
com
Checkpoint 7
Body 1 and body 2 are in a completely inelastic one-dimensional collision.What is
their final momentum if their initial momenta are, respectively, (a) 10 kgm/s and 0;
(b) 10 kgm/s and 4 kgm/s;(c) 10 kgm/s and 4kgm/s?
236 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Additional examples, video, and practice available at WileyPLUS
m
h
M
v
There are two events here.
The bullet collides with the
block. Then the bullet–block
system swings upward by
height h.
Figure 9-17 A ballistic
pendulum, used to
measure the speeds
of bullets.
Sample Problem 9.07 Conservation of momentum, ballistic pendulum
Here is an example of a common technique in physics. We
have a demonstration that cannot be worked out as a whole
(we don’t have a workable equation for it). So, we break it
up into steps that can be worked separately (we have equa-
tions for them).
The ballistic pendulum was used to measure the speeds of
bullets before electronic timing devices were developed. The
version shown in Fig. 9-17 consists of a large block of wood of
mass M5.4 kg, hanging from two long cords. A bullet of
mass m9.5 g is fired into the block, coming quickly to rest.
The block bullet then swing upward, their center of mass
rising a vertical distance h6.3 cm before the pendulum
comes momentarily to rest at the end of its arc. What is the
speed of the bullet just prior to the collision?
KEY IDEAS
We can see that the bullet’s speed vmust determine the rise
height h.However, we cannot use the conservation of mechan-
ical energy to relate these two quantities because surely energy
is transferred from mechanical energy to other forms (such as
thermal energy and energy to break apart the wood) as the
bullet penetrates the block. Nevertheless, we can split this com-
plicated motion into two steps that we can separately analyze:
(1) the bulletblock collision and (2) the bulletblock rise,
during which mechanical energy is conserved.
Reasoning step 1: Because the collision within the
bulletblock system is so brief, we can make two impor-
tant assumptions: (1) During the collision, the gravita-
tional force on the block and the force on the block from
the cords are still balanced. Thus, during the collision, the
net external impulse on the bulletblock system is zero.
Therefore, the system is isolated and its total linear momen-
tum is conserved:
(9-57)
(2) The collision is one-dimensional in the sense that the di-
rection of the bullet and block just after the collision is in the
bullet’s original direction of motion.
Because the collision is one-dimensional, the block is ini-
tially at rest,and the bullet sticks in the block, we use Eq.9-53 to
express the conservation of linear momentum. Replacing the
symbols there with the corresponding symbols here,we have
(9-58)
Reasoning step 2: As the bullet and block now swing up to-
gether, the mechanical energy of the bulletblockEarth
Vm
mMv.
total momentum
before the collision
total momentum
after the collision
.
system is conserved:
(9-59)
(This mechanical energy is not changed by the force of the
cords on the block, because that force is always directed
perpendicular to the block’s direction of travel.) Let’s take the
block’s initial level as our reference level of zero gravitational
potential energy. Then conservation of mechanical energy
means that the system’s kinetic energy at the start of the swing
must equal its gravitational potential energy at the highest
point of the swing. Because the speed of the bullet and block
at the start of the swing is the speed Vimmediately after the
collision,we may write this conservation as
(9-60)
Combining steps: Substituting for Vfrom Eq. 9-58 leads to
(9-61)
(Answer)
The ballistic pendulum is a kind of “transformer, exchang-
ing the high speed of a light object (the bullet) for the low
and thus more easily measurablespeed of a massive ob-
ject (the block).
630 m/s.
0.0095 kg 5.4 kg
0.0095 kg
2(2)(9.8 m/s2)(0.063 m)
vmM
m22gh
1
2(mM)V2(mM)gh.
mechanical energy
at bottom
mechanical energy
at top
.
Elastic Collisions in One Dimension
As we discussed in Module 9-6, everyday collisions are inelastic but we can
approximate some of them as being elastic; that is, we can approximate that the
total kinetic energy of the colliding bodies is conserved and is not transferred to
other forms of energy:
. (9-62)
This means:
total kinetic energy
before the collision
total kinetic energy
after the collision
237
9-7 ELASTIC COLLISIONS IN ONE DIMENSION
9-7 ELASTIC COLLISIONS IN ONE DIMENSION
After reading this module, you should be able to . . .
9.32 For isolated elastic collisions in one dimension, apply the
conservation laws for both the total energy and the net mo-
mentum of the colliding bodies to relate the initial values to
the values after the collision.
9.33 For a projectile hitting a stationary target, identify the re-
sulting motion for the three general cases: equal masses,
target more massive than projectile, projectile more mas-
sive than target.
An elastic collision is a special type of collision in which
the kinetic energy of a system of colliding bodies is con-
served. If the system is closed and isolated, its linear mo-
mentum is also conserved. For a one-dimensional collision in
which body 2 is a target and body 1 is an incoming projec-
tile, conservation of kinetic energy and linear momentum
yield the following expressions for the velocities immediately
after the collision:
and v2f2m1
m1m2
v1i.
v1fm1m2
m1m2
v1i
Learning Objectives
Key Idea
In an elastic collision, the kinetic energy of each colliding body may change, but
the total kinetic energy of the system does not change.
For example, the collision of a cue ball with an object ball in a game of pool
can be approximated as being an elastic collision. If the collision is head-on
(the cue ball heads directly toward the object ball), the kinetic energy of the cue
ball can be transferred almost entirely to the object ball. (Still, the collision trans-
fers some of the energy to the sound you hear.)
Stationary Target
Figure 9-18 shows two bodies before and after they have a one-dimensional colli-
sion, like a head-on collision between pool balls. A projectile body of mass m1
and initial velocity v1imoves toward a target body of mass m2that is initially at
rest (v2i0). Let’s assume that this two-body system is closed and isolated.Then
the net linear momentum of the system is conserved, and from Eq. 9-51 we can write
that conservation as
m1v1im1v1fm2v2f(linear momentum). (9-63)
If the collision is also elastic, then the total kinetic energy is conserved and we
can write that conservation as
(kinetic energy). (9-64)
In each of these equations, the subscript iidentifies the initial velocities and the
subscript fthe final velocities of the bodies. If we know the masses of the bodies
and if we also know v1i, the initial velocity of body 1, the only unknown quantities
are v1fand v2f, the final velocities of the two bodies.With two equations at our dis-
posal, we should be able to find these two unknowns.
1
2m1v1i
21
2m1v1f
21
2m2v2f
2
Figure 9-18 Body 1 moves along an xaxis
before having an elastic collision with
body 2, which is initially at rest. Both
bodies move along that axis after the
collision.
x
Before v1i
m1
Projectile
m2
Target
v2i = 0
x
After
v1f
m1m2
v2f
Here is the generic setup
for an elastic collision with
a stationary target.
To do so, we rewrite Eq. 9-63 as
m1(v1iv1f)m2v2f(9-65)
and Eq. 9-64 as*
(9-66)
After dividing Eq. 9-66 by Eq. 9-65 and doing some more algebra, we obtain
(9-67)
and (9-68)
Note that v2fis always positive (the initially stationary target body with mass m2
always moves forward). From Eq. 9-67 we see that v1fmay be of either sign (the
projectile body with mass m1moves forward if m1m2but rebounds if m1m2).
Let us look at a few special situations.
1. Equal masses If m1m2, Eqs. 9-67 and 9-68 reduce to
v1f0 and v2fv1i,
which we might call a pool player’s result. It predicts that after a head-on colli-
sion of bodies with equal masses, body 1 (initially moving) stops dead in its
tracks and body 2 (initially at rest) takes off with the initial speed of body 1. In
head-on collisions, bodies of equal mass simply exchange velocities. This is
true even if body 2 is not initially at rest.
2. A massive target In Fig. 9-18, a massive target means that m2m1.For
example, we might fire a golf ball at a stationary cannonball. Equations 9-67
and 9-68 then reduce to
(9-69)
This tells us that body 1 (the golf ball) simply bounces back along its incom-
ing path, its speed essentially unchanged. Initially stationary body 2 (the
cannonball) moves forward at a low speed, because the quantity in paren-
theses in Eq. 9-69 is much less than unity.All this is what we should expect.
3. A massive projectile This is the opposite case; that is, m1m2. This time, we
fire a cannonball at a stationary golf ball.Equations 9-67 and 9-68 reduce to
v1fv1iand v2f2v1i. (9-70)
Equation 9-70 tells us that body 1 (the cannonball) simply keeps on going,
scarcely slowed by the collision. Body 2 (the golf ball) charges ahead at twice
the speed of the cannonball. Why twice the speed? Recall the collision de-
scribed by Eq. 9-69, in which the velocity of the incident light body (the golf
ball) changed from vto v, a velocity change of 2v.The same change in ve-
locity (but now from zero to 2v) occurs in this example also.
Moving Target
Now that we have examined the elastic collision of a projectile and a stationary
target, let us examine the situation in which both bodies are moving before they
undergo an elastic collision.
For the situation of Fig. 9-19, the conservation of linear momentum is written as
m1v1im2v2im1v1fm2v2f, (9-71)
v1fv1i and v2f
2m1
m2
v1i.
v2f2m1
m1m2
v1i.
v1fm1m2
m1m2
v1i
m1(v1iv1f)(v1iv1f)m2v2f
2.
238 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
*In this step, we use the identity a2b2(ab)(ab). It reduces the amount of algebra needed to
solve the simultaneous equations Eqs. 9-65 and 9-66.
and the conservation of kinetic energy is written as
(9-72)
To solve these simultaneous equations for v1fand v2f, we first rewrite Eq. 9-71 as
m1(v1iv1f)m2(v2iv2f), (9-73)
and Eq. 9-72 as
m1(v1iv1f)(v1iv1f)m2(v2iv2f)(v2iv2f). (9-74)
After dividing Eq. 9-74 by Eq. 9-73 and doing some more algebra, we obtain
(9-75)
and (9-76)
Note that the assignment of subscripts 1 and 2 to the bodies is arbitrary. If we ex-
change those subscripts in Fig. 9-19 and in Eqs. 9-75 and 9-76, we end up with the
same set of equations. Note also that if we set v2i0, body 2 becomes a stationary
target as in Fig.9-18, and Eqs.9-75 and 9-76 reduce to Eqs. 9-67 and 9-68, respectively.
v2f2m1
m1m2
v1im2m1
m1m2
v2i.
v1fm1m2
m1m2
v1i2m2
m1m2
v2i
1
2m1v1i
21
2m2v2i
21
2m1v1f
21
2m2v2f
2.
239
9-7 ELASTIC COLLISIONS IN ONE DIMENSION
Figure 9-19 Two bodies headed for a one-
dimensional elastic collision.
x
m1
v1i
m2
v2i
Here is the generic setup
for an elastic collision with
a moving target.
Checkpoint 8
What is the final linear momentum of the target in Fig. 9-18 if the initial linear momen-
tum of the projectile is 6 kgm/s and the final linear momentum of the projectile is (a)
2kgm/s and (b) 2kgm/s? (c) What is the final kinetic energy of the target if the
initial and final kinetic energies of the projectile are, respectively, 5 J and 2 J?
two reasons, we can apply Eqs. 9-67 and 9-68 to each of the
collisions.
Calculations: If we start with the first collision, we have too
many unknowns to make any progress: we do not know the
masses or the final velocities of the blocks. So, let’s start with
the second collision in which block 2 stops because of its col-
lision with block 3. Applying Eq. 9-67 to this collision, with
changes in notation, we have
where v2iis the velocity of block 2 just before the collision
and v2fis the velocity just afterward. Substituting v2f0
(block 2 stops) and then m36.0 kg gives us
(Answer)
With similar notation changes, we can rewrite Eq. 9-68 for
the second collision as
where v3fis the final velocity of block 3. Substituting m2m3
and the given v3f5.0 m/s, we find
v2iv3f5.0 m/s.
v3f2m2
m2m3
v2i,
m2m36.00 kg.
v2fm2m3
m2m3
v2i,
Sample Problem 9.08 Chain reaction of elastic collisions
Figure 9-20 Block 1 collides with stationary block 2, which then
collides with stationary block 3.
In Fig. 9-20a, block 1 approaches a line of two stationary
blocks with a velocity of v1i10 m/s. It collides with block 2,
which then collides with block 3, which has mass m36.0 kg.
After the second collision, block 2 is again stationary and
block 3 has velocity v3f5.0 m/s (Fig.9-20b).Assume that the
collisions are elastic. What are the masses of blocks 1 and 2?
What is the final velocity v1fof block 1?
KEY IDEAS
Because we assume that the collisions are elastic, we are to
conserve mechanical energy (thus energy losses to sound,
heating, and oscillations of the blocks are negligible).
Because no external horizontal force acts on the blocks, we
are to conserve linear momentum along the xaxis. For these
(a)
(b)
v1i
v1f
v3f
m1m2m3
x
x
Collisions in Two Dimensions
When two bodies collide, the impulse between them determines the directions in
which they then travel. In particular, when the collision is not head-on, the bodies
do not end up traveling along their initial axis. For such two-dimensional
collisions in a closed, isolated system, the total linear momentum must still be
conserved:
. (9-77)
If the collision is also elastic (a special case), then the total kinetic energy is also
conserved:
K1iK2iK1fK2f. (9-78)
Equation 9-77 is often more useful for analyzing a two-dimensional collision
if we write it in terms of components on an xy coordinate system. For example,
Fig.9-21 shows a glancing collision (it is not head-on) between a projectile body and a
target body initially at rest.The impulses between the bodies have sent the bodies off
at angles u1and u2to the xaxis,along which the projectile initially traveled. In this situ-
P
:
1iP
:
2iP
:
1fP
:
2f
240 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Additional examples, video, and practice available at WileyPLUS
which leads to
(Answer)
Finally, applying Eq. 9-67 to the first collision with this result
and the given v1i, we write
(Answer)
1
3m2m2
1
3m2m2
(10 m/s) 5.0 m/s.
v1fm1m2
m1m2
v1i,
m11
3m21
3(6.0 kg) 2.0 kg.
Next, let’s reconsider the first collision, but we have to
be careful with the notation for block 2: its velocity v2fjust
after the first collision is the same as its velocity v2i(5.0 m/s)
just before the second collision. Applying Eq. 9-68 to the
first collision and using the given v1i10 m/s, we have
5.0 m/s 2m1
m1m2
(10 m/s),
v2f2m1
m1m2
v1i,
9-8 COLLISIONS IN TWO DIMENSIONS
After reading this module, you should be able to . . .
9.34 For an isolated system in which a two-dimensional colli-
sion occurs, apply the conservation of momentum along
each axis of a coordinate system to relate the momentum
components along an axis before the collision to the momen-
tum components along the same axis after the collision.
9.35 For an isolated system in which a two-dimensional elastic
collision occurs, (a) apply the conservation of momentum
along each axis of a coordinate system to relate the momen-
tum components along an axis before the collision to the
momentum components along the same axis after the colli-
sion and (b) apply the conservation of total kinetic energy to
relate the kinetic energies before and after the collision.
If two bodies collide and their motion is not along a single axis
(the collision is not head-on), the collision is two-dimensional.
If the two-body system is closed and isolated, the law of con-
servation of momentum applies to the collision and can be
written as
.P
:
1iP
:
2iP
:
1fP
:
2f
In component form, the law gives two equations that de-
scribe the collision (one equation for each of the two dimen-
sions). If the collision is also elastic (a special case), the
conservation of kinetic energy during the collision gives a
third equation:
K1iK2iK1fK2f.
Learning Objectives
Key Idea
Figure 9-21 An elastic collision between two
bodies in which the collision is not head-
on. The body with mass m2(the target) is
initially at rest.
x
y
θ
2
θ
1 v1i
v2f
v1f
m1
m2
A glancing collision
that conserves
both momentum and
kinetic energy.
ation we would rewrite Eq.9-77 for components along the xaxis as
m1v1im1v1fcos u1m2v2fcos u2, (9-79)
and along the yaxis as
(9-80)
We can also write Eq. 9-78 (for the special case of an elastic collision) in terms of
speeds:
(kinetic energy). (9-81)
Equations 9-79 to 9-81 contain seven variables: two masses, m1and m2; three
speeds, v1i,v1f, and v2f; and two angles, u1and u2. If we know any four of these
quantities, we can solve the three equations for the remaining three quantities.
1
2m1v1i
21
2m1v1f
21
2m2v2f
2
0m1v1fsin u1m2v2fsin u2.
241
9-9 SYSTEMS WITH VARYING MASS: A ROCKET
Checkpoint 9
In Fig. 9-21, suppose that the projectile has an initial momentum of 6 kgm/s, a final
xcomponent of momentum of 4 kgm/s, and a final ycomponent of momentum of
3kgm/s. For the target, what then are (a) the final xcomponent of momentum
and (b) the final ycomponent of momentum?
9-9 SYSTEMS WITH VARYING MASS: A ROCKET
After reading this module, you should be able to . . .
9.36 Apply the first rocket equation to relate the rate at which
the rocket loses mass, the speed of the exhaust products rel-
ative to the rocket, the mass of the rocket, and the accelera-
tion of the rocket.
9.37 Apply the second rocket equation to relate the change in
the rocket’s speed to the relative speed of the exhaust
products and the initial and final mass of the rocket.
9.38 For a moving system undergoing a change in mass at a
given rate, relate that rate to the change in momentum.
In the absence of external forces a rocket accelerates at an
instantaneous rate given by
Rvrel Ma (first rocket equation),
in which Mis the rocket’s instantaneous mass (including
unexpended fuel), Ris the fuel consumption rate, and vrel is
the fuel’s exhaust speed relative to the rocket. The term Rvrel
is the thrust of the rocket engine.
For a rocket with constant Rand vrel, whose speed
changes from vito vfwhen its mass changes from Mito Mf,
(second rocket equation).vfvivrel ln Mi
Mf
Learning Objectives
Key Ideas
Systems with Varying Mass: A Rocket
So far, we have assumed that the total mass of the system remains constant.
Sometimes, as in a rocket, it does not. Most of the mass of a rocket on its launch-
ing pad is fuel, all of which will eventually be burned and ejected from the nozzle
of the rocket engine. We handle the variation of the mass of the rocket as the
rocket accelerates by applying Newton’s second law, not to the rocket alone but
to the rocket and its ejected combustion products taken together.The mass of this
system does not change as the rocket accelerates.
Finding the Acceleration
Assume that we are at rest relative to an inertial reference frame, watching a
rocket accelerate through deep space with no gravitational or atmospheric drag
242 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
Figure 9-22 (a) An accelerating rocket of
mass Mat time t, as seen from an inertial
reference frame. (b) The same but at time
tdt. The exhaust products released dur-
ing interval dt are shown.
forces acting on it. For this one-dimensional motion, let Mbe the mass of the
rocket and vits velocity at an arbitrary time t(see Fig.9-22a).
Figure 9-22bshows how things stand a time interval dt later.The rocket now
has velocity vdv and mass MdM, where the change in mass dM is a negative
quantity. The exhaust products released by the rocket during interval dt have
mass dM and velocity Urelative to our inertial reference frame.
Conserve Momentum. Our system consists of the rocket and the exhaust
products released during interval dt.The system is closed and isolated, so the lin-
ear momentum of the system must be conserved during dt; that is,
PiPf, (9-82)
where the subscripts iand findicate the values at the beginning and end of time
interval dt. We can rewrite Eq. 9-82 as
Mv dM U (MdM)(vdv), (9-83)
where the first term on the right is the linear momentum of the exhaust products
released during interval dt and the second term is the linear momentum of the
rocket at the end of interval dt.
Use Relative Speed. We can simplify Eq. 9-83 by using the relative speed vrel be-
tween the rocket and the exhaust products, which is related to the velocities relative to
the frame with
.
In symbols, this means
(vdv)vrel U,
or Uvdv vrel. (9-84)
Substituting this result for Uinto Eq. 9-83 yields,with a little algebra,
dM vrel M dv. (9-85)
Dividing each side by dt gives us
(9-86)
We replace dM/dt (the rate at which the rocket loses mass) by R, where Ris the
(positive) mass rate of fuel consumption, and we recognize that dv/dt is the accel-
eration of the rocket.With these changes, Eq. 9-86 becomes
Rvrel Ma (first rocket equation). (9-87)
Equation 9-87 holds for the values at any given instant.
Note the left side of Eq. 9-87 has the dimensions of force (kg/sm/s
kgm/s2N) and depends only on design characteristics of the rocket engine
namely, the rate Rat which it consumes fuel mass and the speed vrel with which that
mass is ejected relative to the rocket.We call this term Rvrel the thrust of the rocket
engine and represent it with T.Newton’s second law emerges if we write Eq. 9-87 as
TMa, in which ais the acceleration of the rocket at the time that its mass is M.
Finding the Velocity
How will the velocity of a rocket change as it consumes its fuel? From Eq. 9-85
we have
dv vrel
dM
M.
dM
dt vrel Mdv
dt .
velocity of rocket
relative to frame
velocity of rocket
relative to products
velocity of products
relative to frame
x
v
M
System boundary
(a)
x
v+ dv
M+dM
System boundary
(b)
dM
U
The ejection of mass from
the rocket's rear increases
the rocket's speed.
Integrating leads to
in which Miis the initial mass of the rocket and Mfits final mass. Evaluating the
integrals then gives
(second rocket equation) (9-88)
for the increase in the speed of the rocket during the change in mass from Mito
Mf. (The symbol “ln” in Eq. 9-88 means the natural logarithm.) We see here the
advantage of multistage rockets, in which Mfis reduced by discarding successive
stages when their fuel is depleted. An ideal rocket would reach its destination
with only its payload remaining.
vfvivrel ln Mi
Mf
vf
vi
dv vrel Mf
Mi
dM
M,
243
REVIEW & SUMMARY
rocket’s mass. However, Mdecreases and aincreases as fuel
is consumed. Because we want the initial value of ahere, we
must use the intial value Miof the mass.
Calculation: We find
(Answer)
To be launched from Earth’s surface, a rocket must have
an initial acceleration greater than . That is, it
must be greater than the gravitational acceleration at the
surface. Put another way, the thrust Tof the rocket engine
must exceed the initial gravitational force on the rocket,
which here has the magnitude Mig, which gives us
(850 kg)(9.8 m/s2)8330 N.
Because the acceleration or thrust requirement is not met
(here T6400 N), our rocket could not be launched from
Earth’s surface by itself; it would require another, more
powerful, rocket.
g9.8 m/s2
aT
Mi
6440 N
850 kg 7.6 m/s2.
Sample Problem 9.09 Rocket engine, thrust, acceleration
In all previous examples in this chapter, the mass of a system
is constant (fixed as a certain number). Here is an example of
a system (a rocket) that is losing mass. A rocket whose initial
mass Miis 850 kg consumes fuel at the rate The
speed vrel of the exhaust gases relative to the rocket engine is
2800 m/s.What thrust does the rocket engine provide?
KEY IDEA
Thrust Tis equal to the product of the fuel consumption
rate Rand the relative speed vrel at which exhaust gases are
expelled, as given by Eq. 9-87.
Calculation: Here we find
(Answer)
(b) What is the initial acceleration of the rocket?
KEY IDEA
We can relate the thrust Tof a rocket to the magnitude aof
the resulting acceleration with , where Mis the
TMa
6440 N 6400 N.
TRvrel (2.3 kg/s)(2800 m/s)
R2.3 kg/s.
Additional examples, video, and practice available at WileyPLUS
Center of Mass The center of mass of a system of nparticles is
defined to be the point whose coordinates are given by
(9-5)
or (9-8)
where Mis the total mass of the system.
r
:
com 1
M
n
i1
mir
:
i,
xcom 1
M
n
i1
mixi, ycom 1
M
n
i1
miyi, zcom 1
M
n
i1
mizi,
Review & Summary
Newton’s Second Law for a System of Particles The
motion of the center of mass of any system of particles is governed
by Newton’s second law for a system of particles, which is
. (9-14)
Here is the net force of all the external forces acting on the sys-F
:
net
F
:
net Ma
:
com
tem, Mis the total mass of the system, and is the acceleration
of the system’s center of mass.
a
:com
244 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
must be conserved (it is a constant), which we can write in vector
form as
, (9-50)
where subscripts iand frefer to values just before and just after the
collision, respectively.
If the motion of the bodies is along a single axis,the collision is
one-dimensional and we can write Eq. 9-50 in terms of velocity
components along that axis:
m1v1im2v2im1v1fm2v2f. (9-51)
If the bodies stick together, the collision is a completely
inelastic collision and the bodies have the same final velocity V
(because they are stuck together).
Motion of the Center of Mass The center of mass of a
closed, isolated system of two colliding bodies is not affected by a
collision. In particular, the velocity of the center of mass can-
not be changed by the collision.
Elastic Collisions in One Dimension An elastic collision
is a special type of collision in which the kinetic energy of a system
of colliding bodies is conserved. If the system is closed and
isolated, its linear momentum is also conserved. For a one-
dimensional collision in which body 2 is a target and body 1 is an
incoming projectile, conservation of kinetic energy and linear
momentum yield the following expressions for the velocities
immediately after the collision:
(9-67)
and (9-68)
Collisions in Two Dimensions If two bodies collide and
their motion is not along a single axis (the collision is not head-on),
the collision is two-dimensional. If the two-body system is closed
and isolated, the law of conservation of momentum applies to the
collision and can be written as
. (9-77)
In component form, the law gives two equations that describe the
collision (one equation for each of the two dimensions). If the col-
lision is also elastic (a special case), the conservation of kinetic en-
ergy during the collision gives a third equation:
K1iK2iK1fK2f. (9-78)
Variable-Mass Systems In the absence of external forces a
rocket accelerates at an instantaneous rate given by
Rvrel Ma (first rocket equation), (9-87)
in which Mis the rocket’s instantaneous mass (including
unexpended fuel), Ris the fuel consumption rate,and vrel is the fuel’s
exhaust speed relative to the rocket. The term Rvrel is the thrust of
the rocket engine. For a rocket with constant Rand vrel,whose speed
changes from vito vfwhen its mass changes from Mito Mf,
(second rocket equation). (9-88)vfvivrel ln Mi
Mf
P
:
1iP
:
2iP
:
1fP
:
2f
v2f2m1
m1m2
v1i.
v1fm1m2
m1m2
v1i
v
:com
p
:1ip
:2ip
:1fp
:2f
tum, and is the impulse due to the force exerted on the body
by the other body in the collision:
(9-30)
If Favg is the average magnitude of during the collision and t
is the duration of the collision, then for one-dimensional motion
JFavg t. (9-35)
When a steady stream of bodies, each with mass mand speed v, col-
lides with a body whose position is fixed, the average force on the
fixed body is
(9-37)
where n/tis the rate at which the bodies collide with the fixed
body, and vis the change in velocity of each colliding body. This
average force can also be written as
(9-40)
where m/tis the rate at which mass collides with the fixed body. In
Eqs. 9-37 and 9-40, vvif the bodies stop upon impact and v
2vif they bounce directly backward with no change in their speed.
Conservation of Linear Momentum If a system is isolated
so that no net external force acts on it, the linear momentum of
the system remains constant:
(closed, isolated system). (9-42)
This can also be written as
(closed, isolated system), (9-43)
where the subscripts refer to the values of at some initial time and
at a later time. Equations 9-42 and 9-43 are equivalent statements of
the law of conservation of linear momentum.
Inelastic Collision in One Dimension In an inelastic
collision of two bodies, the kinetic energy of the two-body
system is not conserved (it is not a constant). If the system is
closed and isolated, the total linear momentum of the system
P
:
P
:
iP
:
f
P
:constant
P
:
Favg m
tv,
Favg  n
tp n
tmv,
F
:
(t)
J
:tf
ti
F
:
(t)dt.
F
:
(t)J
:
Linear Momentum and Newton’s Second Law For a sin-
gle particle, we define a quantity called its linear momentum as
, (9-22)
and can write Newton’s second law in terms of this momentum:
(9-23)
For a system of particles these relations become
and (9-25, 9-27)
Collision and Impulse Applying Newton’s second law in
momentum form to a particle-like body involved in a collision
leads to the impulse linear momentum theorem:
, (9-31, 9-32)
where is the change in the body’s linear momen-p
:
fp
:
ip
:
p
:
fp
:
ip
:J
:
F
:
net dP
:
dt .P
:Mv
:
com
F
:
net dp
:
dt .
p
:mv
:
p
:
245
QUESTIONS
Questions
1Figure 9-23 shows an overhead
view of three particles on which ex-
ternal forces act.The magnitudes and
directions of the forces on two of the
particles are indicated. What are the
magnitude and direction of the force
acting on the third particle if the cen-
ter of mass of the three-particle sys-
tem is (a) stationary, (b) moving at a
constant velocity rightward, and (c) accelerating rightward?
2Figure 9-24 shows an over-
head view of four particles of
equal mass sliding over a fric-
tionless surface at constant
velocity. The directions of the
velocities are indicated; their
magnitudes are equal. Consider
pairing the particles. Which
pairs form a system with a cen-
ter of mass that (a) is stationary,
(b) is stationary and at the ori-
gin,and (c) passes through the origin?
3Consider a box that explodes into two pieces while moving with
a constant positive velocity along an xaxis. If one piece, with mass
m1, ends up with positive velocity , then the second piece, with
mass m2, could end up with (a) a positive velocity (Fig. 9-25a), (b)
a negative velocity (Fig. 9-25b), or (c) zero velocity (Fig. 9-25c).
Rank those three possible results for the second piece according to
the corresponding magnitude of , greatest first.v1
:
v2
:
v2
:
v1
:
boxes move over a frictionless confectioner’s counter. For each box,
is its linear momentum conserved along the xaxis and the yaxis?
6Figure 9-28 shows four groups of three or four identical particles
that move parallel to either the xaxis or the yaxis,at identical speeds.
Rank the groups according to center-of-mass speed,greatest first.
y
1
5 N
3 N
2
3
x
Figure 9-23 Question 1.
c d
a
y(m)
2
–2 2 4 –4
–2
x(m)
b
Figure 9-24 Question 2.
v2v2
v1v1
(b) (c)(a)
v1
Figure 9-25 Question 3.
2F0
4F0
6t0
t
F F
t t
F
(a) (b) (c)
3t012t0
2F0
Figure 9-26 Question 4.
xxx
y y y
60°
60°
60°
60° 8 Ν 6 Ν
5 Ν
4 Ν
6 Ν
8 Ν
3 Ν
4 Ν
2 Ν
2 Ν
6 Ν 5 Ν
2 Ν
3 Ν
(a) (b) (c)
Figure 9-27 Question 5.
4Figure 9-26 shows graphs of force magnitude versus time for a
body involved in a collision. Rank the graphs according to the
magnitude of the impulse on the body, greatest first.
5The free-body diagrams in Fig. 9-27 give, from overhead views,
the horizontal forces acting on three boxes of chocolates as the
y
x
(a)
y
x
(c)
y
x
(b)
y
x
(d)
Figure 9-28 Question 6.
7A block slides along a frictionless floor and into a stationary sec-
ond block with the same mass. Figure 9-29 shows four choices for a
graph of the kinetic energies Kof the blocks. (a) Determine which
represent physically impossible situations. Of the others, which best
represents (b) an elastic collision and (c) an inelastic collision?
K
t
(a)
K
t
(b)
K
t
(c)
K
t
(d)
Figure 9-29 Question 7.
8Figure 9-30 shows a snapshot of
block 1 as it slides along an xaxis on a
frictionless floor, before it undergoes
an elastic collision with stationary
block 2.The figure also shows three possible positions of the center of
mass (com) of the two-block system at the time of the snapshot. (Point
Bis halfway between the centers of the two blocks.) Is block 1 station-
ary,moving forward, or moving backward after the collision if the com
is located in the snapshot at (a) A,(b) B,and (c) C?
1 2
A B C
Figure 9-30 Question 8.
9Two bodies have undergone an
elastic one-dimensional collision
along an xaxis. Figure 9-31 is a graph
of position versus time for those
bodies and for their center of mass.
(a) Were both bodies initially moving,
or was one initially stationary? Which
line segment corresponds to the mo-
tion of the center of mass (b) before the collision and (c) after the col-
lision? (d) Is the mass of the body that was moving faster before the
collision greater than,less than, or equal to that of the other body?
10 Figure 9-32: A block on a horizontal floor is initially either
stationary, sliding in the positive direction of an xaxis, or sliding in
246 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
the negative direction of that axis. Then the block explodes into
two pieces that slide along the xaxis. Assume the block and the
two pieces form a closed, isolated system. Six choices for a graph of
the momenta of the block and the pieces are given, all versus time
t. Determine which choices represent physically impossible situa-
tions and explain why.
11 Block 1 with mass m1slides
along an xaxis across a frictionless
floor and then undergoes an elastic
collision with a stationary block 2 with
mass m2. Figure 9-33 shows a plot of
position xversus time tof block 1 until
the collision occurs at position xcand
time tc.In which of the lettered regions
on the graph will the plot be contin-
ued (after the collision) if (a) m1m2
and (b) m1m2? (c) Along which of
the numbered dashed lines will the
plot be continued if m1m2?
12 Figure 9-34 shows four graphs of
position versus time for two bodies
and their center of mass. The two
bodies form a closed, isolated system
and undergo a completely inelastic,
one-dimensional collision on an xaxis.
In graph 1, are (a) the two bodies and
(b) the center of mass moving in the
positive or negative direction of the x
axis? (c) Which of the graphs corre-
spond to a physically impossible situ-
ation? Explain.
x
t
1
2 3
45
6
Figure 9-31 Question 9.
p
t
(a)
p
t
(d)
p
t
(b)
p
t
(e)
p
t
(c)
p
t
(f)
Figure 9-32 Question 10.
x
t
tc
xc
A
B
C
D
54
3
2
1
Figure 9-33 Question 11.
x x
x x
t t
t t
(1) (2)
(3) (4)
Figure 9-34 Question 12.
Module 9-1 Center of Mass
•1 A 2.00 kg particle has the xy coordinates (1.20 m, 0.500 m),
and a 4.00 kg particle has the xy coordinates (0.600 m, 0.750 m).
Both lie on a horizontal plane. At what (a) xand (b) ycoordinates
must you place a 3.00 kg particle such that the center of mass of the
three-particle system has the coor-
dinates (0.500 m, 0.700 m)?
•2 Figure 9-35 shows a three-par-
ticle system, with masses m13.0
kg, m24.0 kg, and m3 8.0 kg.
The scales on the axes are set by
xs2.0 m and ys2.0 m.What are
(a) the xcoordinate and (b) the y
coordinate of the system’s center
of mass? (c) If m3is gradually in-
creased, does the center of mass of the system shift toward or away
from that particle, or does it remain stationary?
••3 Figure 9-36 shows a slab with dimensions d111.0 cm, d2
2.80 cm, and d313.0 cm. Half the slab consists of aluminum (den-
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
y(m)
x(m)
ys
0xs
m1
m3
m2
Figure 9-35 Problem 2.
Aluminum
Iron Midpoint
2d1
d2
d1d1
d3
y
z
x
sity 2.70 g/cm3) and half consists of iron (density 7.85 g/cm3).
What are (a) the xcoordinate, (b) the ycoordinate, and (c) the zco-
ordinate of the slab’s center of mass?
Figure 9-36 Problem 3.
247
PROBLEMS
••4 In Fig. 9-37, three uniform thin rods,
each of length L22 cm, form an in-
verted U. The vertical rods each have a
mass of 14 g; the horizontal rod has a
mass of 42 g. What are (a) the xcoordi-
nate and (b) the ycoordinate of the sys-
tem’s center of mass?
••5 What are (a) the xcoordinate and
(b) the ycoordinate of the center of mass
for the uniform plate shown in Fig. 9-38 if
L5.0 cm?
Module 9-2 Newton’s Second Law for a System of Particles
•9 A stone is dropped at t0.A second stone, with twice the
mass of the first, is dropped from the same point at
t100 ms. (a) How far below the release point is the center of
mass of the two stones at t300 ms? (Neither stone has yet
reached the ground.) (b) How fast is the center of mass of the two-
stone system moving at that time?
•10 A 1000 kg automobile is at rest at a traffic signal. At the in-
stant the light turns green, the automobile starts to move with a
constant acceleration of 4.0 m/s2. At the same instant a 2000 kg
truck, traveling at a constant speed of 8.0 m/s, overtakes and passes
the automobile. (a) How far is the com of the automobiletruck
system from the traffic light at t3.0 s? (b) What is the speed of
the com then?
•11 A big olive (m0.50 kg) lies at the origin of an xy
coordinate system, and a big Brazil nut (M1.5 kg) lies at the
point (1.0, 2.0) m. At t0, a force begins to
act on the olive, and a force begins to act on
the nut. In unit-vector notation, what is the displacement of the
center of mass of the olivenut system at t4.0 s, with respect to
its position at t0?
•12 Two skaters, one with mass 65 kg and the other with mass
40 kg, stand on an ice rink holding a pole of length 10 m and neg-
ligible mass. Starting from the ends of the pole, the skaters pull
themselves along the pole until they meet. How far does the 40
kg skater move?
••13 A shell is shot with an initial velocity of 20 m/s, at
an angle of with the horizontal. At the top of the trajec-
tory, the shell explodes into two fragments of equal mass (Fig.
9-42). One fragment, whose speed immediately after the explo-
sion is zero, falls vertically. How far from the gun does the other
fragment land, assuming that the terrain is level and that air drag
is negligible?
060
v
:
0
SSM
F
:
n(3.0i
ˆ2.0j
ˆ) N
F
:
o(2.0i
ˆ3.0j
ˆ) N
ILW
L
x
y
L
L
Figure 9-37 Problem 4.
3L
4L
2L
2L
2L
4L
L
x
y
Figure 9-38 Problem 5.
••6 Figure 9-39 shows a cubical box that
has been constructed from uniform metal
plate of negligible thickness. The box is
open at the top and has edge length L
40 cm. Find (a) the xcoordinate, (b) the y
coordinate, and (c) the zcoordinate of
the center of mass of the box.
•••7 In the ammonia (NH3) mole-
cule of Fig. 9-40, three hydrogen (H)
atoms form an equilateral triangle, with
the center of the triangle at distance d
9.40 1011 m from each hydrogen
atom. The nitrogen (N) atom is at the
apex of a pyramid, with the three hydro-
gen atoms forming the base. The nitro-
gen-to-hydrogen atomic mass ratio is
13.9, and the nitrogen-to-hydrogen dis-
tance is L10.14 1011 m. What are
the (a) xand (b) ycoordinates of the
molecule’s center of mass?
•••8 A uniform soda can of mass
0.140 kg is 12.0 cm tall and filled with
0.354 kg of soda (Fig. 9-41). Then small
holes are drilled in the top and bottom
(with negligible loss of metal) to drain
the soda. What is the height hof the
com of the can and contents (a) initially
and (b) after the can loses all the soda?
(c) What happens to has the soda
drains out? (d) If xis the height of the
remaining soda at any given instant,
find xwhen the com reaches its lowest
point.
ILW
LOy
x
z
Figure 9-39 Problem 6.
N
L
H
H
H
d
x
y
Figure 9-40 Problem 7.
x
Figure 9-41 Problem 8.
v0
0
Explosion
θ
Figure 9-42 Problem 13.
••14 In Figure 9-43, two particles are launched from the origin of
the coordinate system at time t0. Particle 1 of mass m15.00 g is
shot directly along the xaxis on a frictionless floor, with constant
speed 10.0 m/s. Particle 2 of mass m23.00 g is shot with a velocity
of magnitude 20.0 m/s, at an upward angle such that it always stays
directly above particle 1. (a) What is the maximum height Hmax
reached by the com of the two-particle system? In unit-vector no-
tation, what are the (b) velocity and (c) acceleration of the com
when the com reaches Hmax?
Figure 9-43 Problem 14.
x
y
1
2
248 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
••15 Figure 9-44 shows an arrangement with an air track, in which
a cart is connected by a cord to a hanging block. The cart has mass
m10.600 kg, and its center is initially at xy coordinates (0.500
m, 0 m); the block has mass m20.400 kg,and its center is initially at
xy coordinates (0, 0.100 m).The mass of the cord and pulley are neg-
ligible. The cart is released from rest, and both cart and block move
until the cart hits the pulley. The friction between the cart and the air
track and between the pulley and its axle is negligible. (a) In unit-vec-
tor notation, what is the acceleration of the center of mass of the
cartblock system? (b) What is the velocity of the com as a function
of time t? (c) Sketch the path taken by the com. (d) If the path is
curved, determine whether it bulges upward to the right or downward
to the left, and if it is straight, find the angle between it and the xaxis.
••21 A 0.30 kg softball has a velocity of 15 m/s at an angle of 35be-
low the horizontal just before making contact with the bat.What is the
magnitude of the change in momentum of the ball while in contact
with the bat if the ball leaves with a velocity of (a) 20 m/s, vertically
downward,and (b) 20 m/s,horizontally back toward the pitcher?
••22 Figure 9-47 gives an overhead
view of the path taken by a 0.165 kg
cue ball as it bounces from a rail of a
pool table. The ball’s initial speed is
2.00 m/s, and the angle u1is 30.0.The
bounce reverses the ycomponent of
the ball’s velocity but does not alter
the xcomponent.What are (a) angle
u2and (b) the change in the ball’s lin-
ear momentum in unit-vector nota-
tion? (The fact that the ball rolls is ir-
relevant to the problem.)
Module 9-4 Collision and Impulse
•23 Until his seventies, Henri LaMothe (Fig. 9-48) excited
audiences by belly-flopping from a height of 12 m into 30 cm of
water.Assuming that he stops just as he reaches the bottom of the
water and estimating his mass, find the magnitude of the impulse
on him from the water.
Figure 9-44 Problem 15.
y
x
m2
m1
•••16 Ricardo, of mass 80 kg, and Carmelita, who is lighter,
are enjoying Lake Merced at dusk in a 30 kg canoe.When the ca-
noe is at rest in the placid water, they exchange seats, which are
3.0 m apart and symmetrically lo-
cated with respect to the canoe’s
center. If the canoe moves 40 cm
horizontally relative to a pier post,
what is Carmelita’s mass?
•••17 In Fig. 9-45a, a 4.5 kg dog
stands on an 18 kg flatboat at dis-
tance D6.1 m from the shore. It
walks 2.4 m along the boat toward
shore and then stops. Assuming no
friction between the boat and the wa-
ter, find how far the dog is then from
the shore.(Hint: See Fig.9-45b.)
Module 9-3 Linear Momentum
•18 A 0.70 kg ball moving horizontally at 5.0 m/s strikes a vertical
wall and rebounds with speed 2.0 m/s. What is the magnitude of the
change in its linear momentum?
•19 A 2100 kg truck traveling north at 41 km/h turns east
and accelerates to 51 km/h. (a) What is the change in the truck’s
kinetic energy? What are the (b) magnitude and (c) direction of
the change in its momentum?
••20 At time t0, a ball is
struck at ground level and sent over
level ground.The momentum pver-
sus tduring the flight is given by Fig.
9-46 (with and
). At what initial
angle is the ball launched? (Hint:
Find a solution that does not
require you to read the time of the
low point of the plot.)
p14.0 kgm/s
p06.0 kgm/s
ILW
Dog's displacement dd
Boat's displacement db
(b)
D
(a)
Figure 9-45 Problem 17.
p (kg m/s)
p0
p1531 4 2
t (s)
0
Figure 9-46 Problem 20.
y
x
θ
2
θ
1
Figure 9-47 Problem 22.
George Long/Getty Images, Inc.
Figure 9-48 Problem 23. Belly-flopping into 30 cm of water.
•24 In February 1955, a paratrooper fell 370 m from an air-
plane without being able to open his chute but happened to land in
snow, suffering only minor injuries. Assume that his speed at im-
pact was 56 m/s (terminal speed), that his mass (including gear)
was 85 kg, and that the magnitude of the force on him from the
249
PROBLEMS
12.0 m/s and angle u135.0. Just
after, it is traveling directly upward
with velocity of magnitude 10.0
m/s. The duration of the collision is
2.00 ms. What are the (a) magni-
v
:
2
snow was at the survivable limit of 1.2 105N. What are (a) the
minimum depth of snow that would have stopped him safely and
(b) the magnitude of the impulse on him from the snow?
•25 A 1.2 kg ball drops vertically onto a floor, hitting with a
speed of 25 m/s. It rebounds with an initial speed of 10 m/s. (a)
What impulse acts on the ball during the contact? (b) If the ball is
in contact with the floor for 0.020 s, what is the magnitude of the
average force on the floor from the ball?
•26 In a common but dangerous prank, a chair is pulled away as
a person is moving downward to sit on it, causing the victim to land
hard on the floor. Suppose the victim falls by 0.50 m, the mass that
moves downward is 70 kg,and the collision on the floor lasts 0.082 s.
What are the magnitudes of the (a) impulse and (b) average force
acting on the victim from the floor during the collision?
•27 A force in the negative direction of an xaxis is applied
for 27 ms to a 0.40 kg ball initially moving at 14 m/s in the positive
direction of the axis. The force varies in magnitude, and the im-
pulse has magnitude 32.4 Ns.What are the ball’s (a) speed and (b)
direction of travel just after the force is applied? What are (c) the
average magnitude of the force and (d) the direction of the im-
pulse on the ball?
•28 In tae-kwon-do, a hand is slammed down onto a target
at a speed of 13 m/s and comes to a stop during the 5.0 ms collision.
Assume that during the impact the hand is independent of the arm
and has a mass of 0.70 kg. What are the magnitudes of the (a) im-
pulse and (b) average force on the hand from the target?
•29 Suppose a gangster sprays Superman’s chest with 3 g bullets
at the rate of 100 bullets/min, and the speed of each bullet is 500
m/s. Suppose too that the bullets rebound straight back with no
change in speed. What is the magnitude of the average force on
Superman’s chest?
••30 Two average forces. A steady stream of 0.250 kg snowballs is
shot perpendicularly into a wall at a speed of 4.00 m/s. Each ball
sticks to the wall. Figure 9-49 gives the magnitude Fof the force on
the wall as a function of time tfor two of the snowball impacts.
Impacts occur with a repetition time interval tr50.0 ms, last a du-
ration time interval td10 ms, and produce isosceles triangles on
the graph, with each impact reaching a force maximum Fmax 200 N.
During each impact, what are the magnitudes of (a) the impulse and
(b) the average force on the wall? (c) During a time interval of many
impacts,what is the magnitude of the average force on the wall?
SSM
are the magnitudes of the (c) impulse
and (d) average force (assuming the
same stopping time)?
••32 A 5.0 kg toy car can move
along an xaxis; Fig. 9-50 gives Fxof
the force acting on the car, which be-
gins at rest at time t0. The scale on
the Fxaxis is set by In
unit-vector notation, what is at (a)
t4.0 s and (b) t7.0 s, and (c)
what is at t9.0 s?
••33 Figure 9-51 shows a 0.300
kg baseball just before and just after
it collides with a bat. Just before, the
ball has velocity of magnitudev
:
1
v
:
p
:
Fxs 5.0 N.
Δtr
ΔtdΔtd
F
Fmax
t
Figure 9-49 Problem 30.
24 86
Fxs
Fxs
t (s)
Fx(N)
Figure 9-50 Problem 32.
θ
v2v1
1
y
x
Figure 9-51 Problem 33.
Stephen Dalton/Photo Researchers, Inc.
Figure 9-52 Problem 34. Lizard running across water.
••31 Jumping up before the elevator hits. After the cable
snaps and the safety system fails, an elevator cab free-falls from a
height of 36 m. During the collision at the bottom of the elevator
shaft, a 90 kg passenger is stopped in 5.0 ms. (Assume that neither the
passenger nor the cab rebounds.) What are the magnitudes of the (a)
impulse and (b) average force on the passenger during the collision?
If the passenger were to jump upward with a speed of 7.0 m/s relative
to the cab floor just before the cab hits the bottom of the shaft, what
tude and (b) direction (relative to the positive direction of the x
axis) of the impulse on the ball from the bat? What are the (c)
magnitude and (d) direction of the average force on the ball from
the bat?
••34 Basilisk lizards can run across the top of a water sur-
face (Fig. 9-52). With each step, a lizard first slaps its foot against
the water and then pushes it down into the water rapidly enough to
form an air cavity around the top of the foot. To avoid having to
pull the foot back up against water drag in order to complete the
step, the lizard withdraws the foot before water can flow into the
air cavity. If the lizard is not to sink, the average upward impulse
on the lizard during this full action of slap, downward push, and
withdrawal must match the downward impulse due to the gravita-
tional force. Suppose the mass of a basilisk lizard is 90.0 g, the mass
of each foot is 3.00 g, the speed of a foot as it slaps the water is
1.50 m/s, and the time for a single step is 0.600 s. (a) What is the
magnitude of the impulse on the lizard during the slap? (Assume
this impulse is directly upward.) (b) During the 0.600 s duration of
a step, what is the downward impulse on the lizard due to the gravi-
tational force? (c) Which action, the slap or the push, provides the
primary support for the lizard, or are they approximately equal in
their support?
250 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
μ
= 0
μ
L
μ
R
dR
dL
Figure 9-57 Problem 44.
••35 Figure 9-53 shows an
approximate plot of force mag-
nitude Fversus time tduring the
collision of a 58 g Superball with
a wall. The initial velocity of the
ball is 34 m/s perpendicular to
the wall; the ball rebounds di-
rectly back with approximately
the same speed, also perpendi-
cular to the wall. What is Fmax,
the maximum magnitude of the
force on the ball from the wall during the collision?
••36 A 0.25 kg puck is initially stationary on an ice surface with
negligible friction. At time t0, a horizontal force begins to
move the puck.The force is given by (12.0 , with
in newtons and tin seconds, and it acts until its magnitude is
zero. (a) What is the magnitude of the impulse on the puck from
the force between t0.500 s and t1.25 s? (b) What is the
change in momentum of the puck between t0 and the instant
at which F0?
••37 A soccer player kicks a soccer ball of mass 0.45 kg that
is initially at rest. The foot of the player is in contact with the ball
for 3.0 103s,and the force of the kick is given by
F(t)[(6.0 106)t(2.0 109)t2] N
for 0 t3.0 103s, where tis in seconds. Find the magnitudes
of (a) the impulse on the ball due to the kick, (b) the average force
on the ball from the player’s foot during the period of contact,
(c) the maximum force on the ball from the player’s foot during the
period of contact, and (d) the ball’s velocity immediately after it
loses contact with the player’s foot.
••38 In the overhead view of Fig.
9-54, a 300 g ball with a speed vof
6.0 m/s strikes a wall at an angle u
of 30and then rebounds with the
same speed and angle. It is in con-
tact with the wall for 10 ms. In unit-
vector notation, what are (a) the
impulse on the ball from the wall
and (b) the average force on the wall from the ball?
SSM
F
:
3.00t2)i
ˆ
F
:
locity that the explosion gives the rest of the rocket. (2) Next, at
time t0.80 s, block Ris shot to the right with a speed of 3.00 m/s
relative to the velocity that block Cthen has. At t2.80 s, what
are (a) the velocity of block Cand (b) the position of its center?
••42 An object, with mass mand speed vrelative to an observer,
explodes into two pieces, one three times as massive as the other;
the explosion takes place in deep space. The less massive piece
stops relative to the observer. How much kinetic energy is added
to the system during the explosion, as measured in the observer’s
reference frame?
••43 In the Olympiad of 708 B.C., some athletes competing in
the standing long jump used handheld weights called halteres to
lengthen their jumps (Fig.9-56).The weights were swung up in front
just before liftoff and then swung down and thrown backward dur-
ing the flight. Suppose a modern 78 kg long jumper similarly uses
two 5.50 kg halteres, throwing them horizontally to the rear at his
maximum height such that their horizontal velocity is zero rela-
tive to the ground. Let his liftoff velocity be m/s
with or without the halteres, and assume that he lands at the liftoff
level. What distance would the use of the halteres add to his range?
v
: (9.5i
ˆ4.0j
ˆ)
Module 9-5 Conservation of Linear Momentum
•39 A 91 kg man lying on a surface of negligible friction
shoves a 68 g stone away from himself, giving it a speed of 4.0 m/s.
What speed does the man acquire as a result?
•40 A space vehicle is traveling at 4300 km/h relative to Earth
when the exhausted rocket motor (mass 4m) is disengaged and
sent backward with a speed of 82 km/h relative to the command
module (mass m).What is the speed of the command module rel-
ative to Earth just after the separation?
••41 Figure 9-55 shows a two-ended “rocket” that is initially sta-
tionary on a frictionless floor, with its center at the origin of an x
axis.The rocket consists of a central block C(of mass M6.00 kg)
and blocks Land R(each of mass m2.00 kg) on the left and
right sides. Small explosions can
shoot either of the side blocks away
from block Cand along the xaxis.
Here is the sequence: (1) At time t
0, block Lis shot to the left with a
speed of 3.00 m/s relative to the ve-
SSM
y
x
θ θ
v
v
Figure 9-54 Problem 38.
x
L R
C
0
Figure 9-55 Problem 41.
••44 In Fig. 9-57, a stationary block explodes into two pieces L
and Rthat slide across a frictionless floor and then into regions with
friction, where they stop. Piece L, with a mass of 2.0 kg, encounters a
coefficient of kinetic friction mL0.40 and slides to a stop in distance
dL0.15 m. Piece Rencounters a coefficient of kinetic friction mR
0.50 and slides to a stop in distance dR0.25 m. What was the mass
of the block?
Réunion des Musées Nationaux/
Art Resource
Figure 9-56 Problem 43.
24
Fmax
t(ms)
F (N)
60
0
Figure 9-53 Problem 35.
••45 A 20.0 kg body is moving through space in the
positive direction of an xaxis with a speed of 200 m/s when, due
to an internal explosion, it breaks into three parts. One part, with a
mass of 10.0 kg, moves away from the point of explosion with
a speed of 100 m/s in the positive ydirection.A second part, with a
mass of 4.00 kg, moves in the negative xdirection with a speed of
500 m/s. (a) In unit-vector notation, what is the velocity of the third
part? (b) How much energy is released in the explosion? Ignore ef-
fects due to the gravitational force.
••46 A 4.0 kg mess kit sliding on a frictionless surface explodes
into two 2.0 kg parts: 3.0 m/s, due north, and 5.0 m/s, 30north of
east.What is the original speed of the mess kit?
WWWSSM
251
PROBLEMS
••47 A vessel at rest at the origin of an xy coordinate system ex-
plodes into three pieces. Just after the explosion, one piece, of mass
m, moves with velocity (30 m/s) and a second piece, also of mass
m, moves with velocity (30 m/s) . The third piece has mass 3m.
Just after the explosion, what are the (a) magnitude and (b) direc-
tion of the velocity of the third piece?
•••48 Particle Aand particle Bare held together with a com-
pressed spring between them. When they are released, the spring
pushes them apart, and they then fly off in opposite directions, free of
the spring.The mass of Ais 2.00 times the mass of B, and the energy
stored in the spring was 60 J.Assume that the spring has negligible
mass and that all its stored energy is transferred to the particles.
Once that transfer is complete, what are the kinetic energies of (a)
particle Aand (b) particle B?
Module 9-6 Momentum and Kinetic Energy in Collisions
•49 A bullet of mass 10 g strikes a ballistic pendulum of mass
2.0 kg.The center of mass of the pendulum rises a vertical distance
of 12 cm. Assuming that the bullet remains embedded in the pen-
dulum, calculate the bullet’s initial speed.
•50 A 5.20 g bullet moving at 672 m/s strikes a 700 g wooden
block at rest on a frictionless surface.The bullet emerges, traveling
in the same direction with its speed reduced to 428 m/s. (a) What is
the resulting speed of the block? (b) What is the speed of the
bulletblock center of mass?
••51 In Fig. 9-58a, a 3.50 g bullet is fired horizontally at two
blocks at rest on a frictionless table.The bullet passes through block
1 (mass 1.20 kg) and embeds itself in block 2 (mass 1.80 kg). The
blocks end up with speeds v10.630 m/s and v21.40 m/s (Fig.
9-58b). Neglecting the material removed from block 1 by the bullet,
find the speed of the bullet as it (a) leaves and (b) enters block 1.
j
ˆ
i
ˆ
collisions (CVC). (b) What percent of the original kinetic energy is
lost if the car hits a 300 kg camel? (c) Generally, does the percent
loss increase or decrease if the animal mass decreases?
••54 A completely inelastic collision occurs between two balls of
wet putty that move directly toward each other along a vertical
axis. Just before the collision, one ball, of mass 3.0 kg, is moving up-
ward at 20 m/s and the other ball, of mass 2.0 kg, is moving down-
ward at 12 m/s. How high do the combined two balls of putty rise
above the collision point? (Neglect air drag.)
••55 A 5.0 kg block with a speed of 3.0 m/s collides with a 10
kg block that has a speed of 2.0 m/s in the same direction.After the
collision, the 10 kg block travels in the original direction with a
speed of 2.5 m/s. (a) What is the velocity of the 5.0 kg block imme-
diately after the collision? (b) By how much does the total kinetic
energy of the system of two blocks change because of the colli-
sion? (c) Suppose, instead, that the 10 kg block ends up with a
speed of 4.0 m/s. What then is the change in the total kinetic en-
ergy? (d) Account for the result you obtained in (c).
••56 In the “before” part of Fig. 9-60, car A(mass 1100 kg) is
stopped at a traffic light when it is rear-ended by car B(mass
1400 kg). Both cars then slide with locked wheels until the fric-
tional force from the slick road (with a low mkof 0.13) stops them,
at distances dA8.2 m and dB6.1 m.What are the speeds of (a)
car Aand (b) car Bat the start of the sliding, just after the colli-
sion? (c) Assuming that linear momentum is conserved during
the collision, find the speed of car Bjust before the collision.
(d) Explain why this assumption may be invalid.
ILW
••52 In Fig. 9-59, a 10 g bullet
moving directly upward at 1000 m/s
strikes and passes through the cen-
ter of mass of a 5.0 kg block initially
at rest. The bullet emerges from the
block moving directly upward at 400
m/s. To what maximum height does
the block then rise above its initial
position?
••53 In Anchorage, collisions of a vehicle with a moose are so
common that they are referred to with the abbreviation MVC.
Suppose a 1000 kg car slides into a stationary 500 kg moose on a
very slippery road, with the moose being thrown through the wind-
shield (a common MVC result). (a) What percent of the original
kinetic energy is lost in the collision to other forms of energy? A
similar danger occurs in Saudi Arabia because of camel–vehicle
1 2
Frictionless
(a)
(b)
v1v2
Figure 9-58 Problem 51.
Bullet
Figure 9-59 Problem 52.
••57 In Fig. 9-61, a ball of mass
Figure 9-60 Problem 56.
AB
AB
dA
dB
Before
After
v0
vim
M
Figure 9-61 Problem 57.
m60 g is shot with speed vi22
m/s into the barrel of a spring gun of
mass M240 g initially at rest on a
frictionless surface.The ball sticks in
the barrel at the point of maximum compression of the spring.
Assume that the increase in thermal energy due to friction be-
tween the ball and the barrel is negligible. (a) What is the speed of
the spring gun after the ball stops in the barrel? (b) What fraction
of the initial kinetic energy of the ball is stored in the spring?
•••58 In Fig. 9-62, block 2 (mass 1.0
kg) is at rest on a frictionless surface
and touching the end of an un-
stretched spring of spring constant
200 N/m.The other end of the spring
is fixed to a wall. Block 1 (mass 2.0 kg), traveling at speed v14.0
m/s,collides with block 2, and the two blocks stick together.When the
blocks momentarily stop, by what distance is the spring compressed?
1 2
v1
Figure 9-62 Problem 58.
252 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
h
1
2
Frictionless
μ
k
Figure 9-67 Problem 68.
Basketball
Baseball
(a) Before (b) After
Figure 9-68 Problem 69.
1 2
2d d
Figure 9-69 Problem 70.
•••59 In Fig. 9-63, block 1 (mass 2.0 kg) is moving rightward at
10 m/s and block 2 (mass 5.0 kg) is moving rightward at 3.0 m/s.
The surface is frictionless, and a spring with a spring constant of
1120 N/m is fixed to block 2.When the blocks collide, the compres-
sion of the spring is maximum at the instant the blocks have the
same velocity. Find the maximum compression.
ILW friction is 0.50; there they stop. How far into that region do (a)
block 1 and (b) block 2 slide?
••67 In Fig. 9-66, particle 1 of mass
m10.30 kg slides rightward along
an xaxis on a frictionless floor with a
speed of 2.0 m/s.When it reaches x
0, it undergoes a one-dimensional
elastic collision with stationary parti-
cle 2 of mass m20.40 kg.When par-
ticle 2 then reaches a wall at xw70 cm, it bounces from the wall
with no loss of speed. At what position on the xaxis does particle 2
then collide with particle 1?
••68 In Fig. 9-67, block 1 of mass m1slides from rest along a
frictionless ramp from height h2.50 m and then collides with
stationary block 2, which has mass m22.00m1.After the collision,
block 2 slides into a region where the coefficient of kinetic friction
mkis 0.500 and comes to a stop in distance dwithin that region.
What is the value of distance dif the collision is (a) elastic and (b)
completely inelastic?
Module 9-7 Elastic Collisions in One Dimension
•60 In Fig. 9-64, block A(mass 1.6
kg) slides into block B(mass 2.4 kg),
along a frictionless surface. The direc-
tions of three velocities before (i) and
after ( f) the collision are indicated;
the corresponding speeds are vAi
5.5 m/s, vBi 2.5 m/s, and vBf 4.9
m/s. What are the (a) speed and (b)
direction (left or right) of velocity
? (c) Is the collision elastic?
•61 A cart with mass 340 g
moving on a frictionless linear air track at an initial speed of 1.2 m/s
undergoes an elastic collision with an initially stationary cart of un-
known mass.After the collision, the first cart continues in its origi-
nal direction at 0.66 m/s. (a) What is the mass of the second cart?
(b) What is its speed after impact? (c) What is the speed of the two-
cart center of mass?
•62 Two titanium spheres approach each other head-on with the
same speed and collide elastically. After the collision, one of the
spheres, whose mass is 300 g, remains at rest. (a) What is the mass
of the other sphere? (b) What is the speed of the two-sphere center
of mass if the initial speed of each sphere is 2.00 m/s?
••63 Block 1 of mass m1slides along a frictionless floor and into a
one-dimensional elastic collision with stationary block 2 of mass
m23m1. Prior to the collision, the center of mass of the two-
block system had a speed of 3.00 m/s. Afterward, what are the
speeds of (a) the center of mass and (b) block 2?
••64 A steel ball of mass 0.500 kg
is fastened to a cord that is 70.0 cm long
and fixed at the far end.The ball is then
released when the cord is horizontal
(Fig. 9-65). At the bottom of its path,
the ball strikes a 2.50 kg steel block ini-
tially at rest on a frictionless surface.
The collision is elastic. Find (a) the
speed of the ball and (b) the speed of
the block, both just after the collision.
••65 A body of mass 2.0 kg makes an elastic collision with
another body at rest and continues to move in the original
direction but with one-fourth of its original speed. (a) What is the
mass of the other body? (b) What is the speed of the two-body cen-
ter of mass if the initial speed of the 2.0 kg body was 4.0 m/s?
••66 Block 1, with mass m1and speed 4.0 m/s, slides along an x
axis on a frictionless floor and then undergoes a one-dimensional
elastic collision with stationary block 2, with mass m20.40m1.The
two blocks then slide into a region where the coefficient of kinetic
SSM
SSM
v
:
Af
1 2
Figure 9-63 Problem 59.
vAf = ? vBf
vAi vBi
Figure 9-64 Problem 60.
Figure 9-65 Problem 64.
x (cm)
0xw
1 2
Figure 9-66 Problem 67.
•••69 A small ball of
mass mis aligned above a larger ball
of mass M0.63 kg (with a slight
separation, as with the baseball and
basketball of Fig. 9-68a), and the
two are dropped simultaneously
from a height of h1.8 m.
(Assume the radius of each ball is
negligible relative to h.) (a) If the
larger ball rebounds elastically
from the floor and then the small
ball rebounds elastically from the
larger ball, what value of mresults
in the larger ball stopping when it
collides with the small ball? (b)
What height does the small ball
then reach (Fig. 9-68b)?
•••70 In Fig. 9-69, puck 1 of mass m10.20 kg is sent sliding
across a frictionless lab bench, to undergo a one-dimensional elas-
tic collision with stationary puck 2. Puck 2 then slides off the bench
and lands a distance dfrom the base of the bench. Puck 1 rebounds
from the collision and slides off the opposite edge of the bench,
landing a distance 2dfrom the base of the bench. What is the mass
of puck 2? (Hint: Be careful with signs.)
253
PROBLEMS
Module 9-8 Collisions in Two Dimensions
••71 In Fig. 9-21, projectile particle 1 is an alpha particle and
target particle 2 is an oxygen nucleus. The alpha particle is scattered
at angle u164.0and the oxygen nucleus recoils with speed 1.20
105m/s and at angle u251.0. In atomic mass units, the mass of the
alpha particle is 4.00 u and the mass of the oxygen nucleus is 16.0 u.
What are the (a) final and (b) initial speeds of the alpha particle?
••72 Ball B, moving in the positive direction of an xaxis at speed
v, collides with stationary ball Aat the origin. Aand Bhave differ-
ent masses.After the collision, Bmoves in the negative direction of
the yaxis at speed v/2. (a) In what direction does Amove?
(b) Show that the speed of Acannot be determined from the given
information.
••73 After a completely inelastic collision, two objects of the same
mass and same initial speed move away together at half their initial
speed. Find the angle between the initial velocities of the objects.
••74 Two 2.0 kg bodies, Aand B, collide. The velocities before the
collision are and m/s.After
the collision, What are (a) the final velocity
of Band (b) the change in the total kinetic energy (including sign)?
••75 A projectile proton with a speed of 500 m/s collides elasti-
cally with a target proton initially at rest. The two protons then
move along perpendicular paths, with the projectile path at 60
from the original direction.After the collision, what are the speeds
of (a) the target proton and (b) the projectile proton?
Module 9-9 Systems with Varying Mass: A Rocket
•76 A 6090 kg space probe moving nose-first toward Jupiter at
105 m/s relative to the Sun fires its rocket engine, ejecting 80.0 kg
of exhaust at a speed of 253 m/s relative to the space probe.What is
the final velocity of the probe?
•77 In Fig. 9-70, two long barges are moving in the same
direction in still water, one with a speed of 10 km/h and the other
with a speed of 20 km/h.While they are passing each other, coal is
shoveled from the slower to the faster one at a rate of 1000 kg/min.
How much additional force must be provided by the driving en-
gines of (a) the faster barge and (b) the slower barge if neither is to
change speed? Assume that the shoveling is always perfectly side-
ways and that the frictional forces between the barges and the water
do not depend on the mass of the barges.
SSM
20j
ˆ) m/s.v
:
A(5.0i
ˆ
(10i
ˆ5.0j
ˆ)v
:
Bv
:
A(15i
ˆ30j
ˆ) m/s
ILW
certain interval. What must be the rocket’s mass ratio (ratio of ini-
tial to final mass) over that interval if the rocket’s original speed
relative to the inertial frame is to be equal to (a) the exhaust speed
(speed of the exhaust products relative to the rocket) and (b) 2.0
times the exhaust speed?
•79 A rocket that is in deep space and initially at rest
relative to an inertial reference frame has a mass of 2.55 105kg,
of which 1.81 105kg is fuel. The rocket engine is then fired for
250 s while fuel is consumed at the rate of 480 kg/s. The speed of
the exhaust products relative to the rocket is 3.27 km/s. (a) What is
the rocket’s thrust? After the 250 s firing, what are (b) the mass
and (c) the speed of the rocket?
Additional Problems
80 An object is tracked by a radar station and determined to have
a position vector given by (3500 160t)2700 300 , with
in meters and tin seconds.The radar station’s xaxis points east,
its yaxis north, and its zaxis vertically up. If the object is a 250 kg
meteorological missile, what are (a) its linear momentum, (b) its
direction of motion, and (c) the net force on it?
81 The last stage of a rocket, which is traveling at a speed of
7600 m/s, consists of two parts that are clamped together: a rocket
case with a mass of 290.0 kg and a payload capsule with a mass of
150.0 kg. When the clamp is released, a compressed spring causes
the two parts to separate with a relative speed of 910.0 m/s. What
are the speeds of (a) the rocket case and (b) the payload after they
have separated? Assume that all velocities are along the same line.
Find the total kinetic energy of the two parts (c) before and (d) after
they separate. (e) Account for the difference.
82 Pancake collapse of a tall
building. In the section of a tall
building shown in Fig. 9-71a, the in-
frastructure of any given floor K
must support the weight Wof all
higher floors. Normally the infra-
structure is constructed with a
safety factor sso that it can with-
stand an even greater downward
force of sW. If, however, the support
columns between Kand Lsuddenly
collapse and allow the higher floors to free-fall together onto floor
K(Fig. 9-71b), the force in the collision can exceed sW and, after a
brief pause, cause Kto collapse onto floor J, which collapses on
floor I, and so on until the ground is reached. Assume that the
floors are separated by and have the same mass.Also as-
sume that when the floors above Kfree-fall onto K, the collision
lasts 1.5 ms. Under these simplified conditions, what value must the
safety factor sexceed to prevent pancake collapse of the building?
83 “Relative” is an important
word. In Fig. 9-72, block Lof mass
mL1.00 kg and block Rof mass
mR0.500 kg are held in place with
a compressed spring between them.
When the blocks are released, the spring sends them sliding across
a frictionless floor. (The spring has negligible mass and falls to the
floor after the blocks leave it.) (a) If the spring gives block La re-
lease speed of 1.20 m/s relative to the floor, how far does block R
travel in the next 0.800 s? (b) If, instead, the spring gives block La
release speed of 1.20 m/s relative to the velocity that the spring
gives block R, how far does block Rtravel in the next 0.800 s?
d4.0 m
r
:
k
ˆ
j
ˆ
i
ˆ
r
:
ILWSSM
Figure 9-70 Problem 77.
N
M
L
K
J
I
d
(a) (b)
Figure 9-71 Problem 82.
L R
Figure 9-72 Problem 83.
•78 Consider a rocket that is in deep space and at rest relative to
an inertial reference frame. The rocket’s engine is to be fired for a
254 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
84 Figure 9-73 shows an overhead
view of two particles sliding at constant
velocity over a frictionless surface. The
particles have the same mass and the
same initial speed v4.00 m/s, and they
collide where their paths intersect. An
xaxis is arranged to bisect the angle be-
tween their incoming paths, such that
u40.0. The region to the right of the
collision is divided into four lettered
sections by the xaxis and four numbered dashed lines. In what re-
gion or along what line do the particles travel if the collision is (a)
completely inelastic, (b) elastic, and (c) inelastic? What are their fi-
nal speeds if the collision is (d) completely inelastic and (e) elastic?
85 Speed deamplifier. In Fig.
9-74, block 1 of mass m1slides along
an xaxis on a frictionless floor at
speed 4.00 m/s. Then it undergoes a
one-dimensional elastic collision
with stationary block 2 of mass m2
2.00m1. Next, block 2 undergoes a one-dimensional elastic collision
with stationary block 3 of mass m32.00m2. (a) What then is the
speed of block 3? Are (b) the speed, (c) the kinetic energy, and (d)
the momentum of block 3 greater than, less than, or the same as
the initial values for block 1?
origin with linear momentum (6.4 1023 kgm/s) . What are
the (a) magnitude and (b) direction of the linear momentum of the
daughter nucleus? (c) If the daughter nucleus has a mass of 5.8
1026 kg,what is its kinetic energy?
91 A 75 kg man rides on a 39 kg cart moving at a velocity of 2.3 m/s.
He jumps off with zero horizontal velocity relative to the ground.
What is the resulting change in the cart’s velocity,including sign?
92 Two blocks of masses 1.0 kg and 3.0 kg are connected by a
spring and rest on a frictionless surface. They are given velocities
toward each other such that the 1.0 kg block travels initially at
1.7 m/s toward the center of mass, which remains at rest. What is
the initial speed of the other block?
93 A railroad freight car of mass 3.18 104kg collides
with a stationary caboose car.They couple together, and 27.0% of
the initial kinetic energy is transferred to thermal energy, sound,
vibrations, and so on. Find the mass of the caboose.
94 An old Chrysler with mass 2400 kg is moving along a straight
stretch of road at 80 km/h. It is followed by a Ford with mass 1600
kg moving at 60 km/h. How fast is the center of mass of the two
cars moving?
95 In the arrangement of Fig.9-21, billiard ball 1 moving at a
speed of 2.2 m/s undergoes a glancing collision with identical bil-
liard ball 2 that is at rest. After the collision, ball 2 moves at speed
1.1 m/s, at an angle of u260.What are (a) the magnitude and (b)
the direction of the velocity of ball 1 after the collision? (c) Do the
given data suggest the collision is elastic or inelastic?
96 A rocket is moving away from the solar system at a speed of
6.0 103m/s. It fires its engine, which ejects exhaust with a speed
of 3.0 103m/s relative to the rocket. The mass of the rocket at
this time is 4.0 104kg, and its acceleration is 2.0 m/s2. (a) What is
the thrust of the engine? (b) At what rate, in kilograms per second,
is exhaust ejected during the firing?
SSM
SSM
j
ˆ
A
D
B
C
θθ
θθ
x
43
2
1
Figure 9-73 Problem 84.
x
321
Figure 9-74 Problem 85.
x
321
Figure 9-75 Problem 86.
Figure 9-76 Problem 97.
1
2
3
v0x
The velocity of ball 1 has magnitude v010 m/s and is directed at
the contact point of balls 1 and 2. After the collision, what are the
(a) speed and (b) direction of the velocity of ball 2, the (c) speed
and (d) direction of the velocity of ball 3, and the (e) speed and (f)
direction of the velocity of ball 1? (Hint: With friction absent, each
impulse is directed along the line connecting the centers of the col-
liding balls,normal to the colliding surfaces.)
98 A 0.15 kg ball hits a wall with a velocity of (5.00 m/s) (6.50
m/s) (4.00 m/s) . It rebounds from the wall with a velocity of
(2.00 m/s) (3.50 m/s) ( 3.20 m/s) . What are
(a) the change in the ball’s momentum, (b) the im-
pulse on the ball, and (c) the impulse on the wall?
99 In Fig. 9-77, two identical containers of sugar
are connected by a cord that passes over a friction-
less pulley. The cord and pulley have negligible
mass, each container and its sugar together have a
mass of 500 g, the centers of the containers are sepa-
rated by 50 mm, and the containers are held fixed at
the same height. What is the horizontal distance be-
tween the center of container 1 and the center
of mass of the two-container system (a) initially and
k
ˆ
j
ˆi
ˆ
k
ˆ
j
ˆi
ˆ
97 The three balls in the
overhead view of Fig. 9-76 are
identical. Balls 2 and 3 touch
each other and are aligned per-
pendicular to the path of ball 1.
Figure 9-77
Problem 99.
1 2
sion with stationary block 2 of mass m20.500m1.Next, block 2 un-
dergoes a one-dimensional elastic collision with stationary block 3
of mass m30.500m2.(a) What then is the speed of block 3? Are (b)
the speed, (c) the kinetic energy, and (d) the momentum of block 3
greater than, less than, or the same as the initial values for block 1?
87 A ball having a mass of 150 g strikes a wall with a speed of
5.2 m/s and rebounds with only 50% of its initial kinetic energy. (a)
What is the speed of the ball immediately after rebounding? (b)
What is the magnitude of the impulse on the wall from the ball? (c) If
the ball is in contact with the wall for 7.6 ms, what is the magnitude of
the average force on the ball from the wall during this time interval?
88 A spacecraft is separated into two parts by detonating the ex-
plosive bolts that hold them together. The masses of the parts are
1200 kg and 1800 kg; the magnitude of the impulse on each part
from the bolts is 300 Ns. With what relative speed do the two
parts separate because of the detonation?
89 A 1400 kg car moving at 5.3 m/s is initially traveling
north along the positive direction of a yaxis. After completing a
90right-hand turn in 4.6 s, the inattentive operator drives into a
tree, which stops the car in 350 ms. In unit-vector notation, what is
the impulse on the car (a) due to the turn and (b) due to the colli-
sion? What is the magnitude of the average force that acts on the
car (c) during the turn and (d) during the collision? (e) What is the
direction of the average force during the turn?
90 A certain radioactive (parent) nucleus transforms to a dif-
ferent (daughter) nucleus by emitting an electron and a neutrino.
The parent nucleus was at rest at the origin of an xy coordinate sys-
tem. The electron moves away from the origin with linear momen-
tum (1.2 1022 kgm/s) ; the neutrino moves away from thei
ˆ
ILW
SSM
86 Speed amplifier. In Fig. 9-75,
block 1 of mass m1slides along an x
axis on a frictionless floor with a
speed of v1i4.00 m/s. Then it under-
goes a one-dimensional elastic colli-
255
PROBLEMS
(b) after 20 g of sugar is transferred from container 1 to container
2? After the transfer and after the containers are released, (c) in
what direction and (d) at what acceleration magnitude does the
center of mass move?
100 In a game of pool, the cue ball strikes another ball of the
same mass and initially at rest. After the collision, the cue ball
moves at 3.50 m/s along a line making an angle of 22.0with the
cue ball’s original direction of motion, and the second ball has a
speed of 2.00 m/s. Find (a) the angle between the direction of mo-
tion of the second ball and the original direction of motion of the
cue ball and (b) the original speed of the cue ball. (c) Is kinetic en-
ergy (of the centers of mass, don’t consider the rotation) con-
served?
101 In Fig. 9-78, a 3.2 kg box of
running shoes slides on a horizontal
frictionless table and collides with a
2.0 kg box of ballet slippers initially
at rest on the edge of the table, at
height h0.40 m. The speed of the
3.2 kg box is 3.0 m/s just before the
collision. If the two boxes stick to-
gether because of packing tape on their
sides, what is their kinetic energy just before
they strike the floor?
102 In Fig. 9-79, an 80 kg man is on a lad-
der hanging from a balloon that has a total
mass of 320 kg (including the basket passen-
ger). The balloon is initially stationary rela-
tive to the ground. If the man on the ladder
begins to climb at 2.5 m/s relative to the lad-
der, (a) in what direction and (b) at what
speed does the balloon move? (c) If the man
then stops climbing, what is the speed of the
balloon?
103 In Fig. 9-80, block 1 of mass m16.6 kg
is at rest on a long frictionless table that is up
against a wall. Block 2 of mass m2is placed
between block 1 and the wall and sent sliding
to the left, toward block 1, with constant
speed v2i. Find the value of m2for which both
blocks move with the same velocity after block 2 has collided once
with block 1 and once with the wall.Assume all collisions are elastic
(the collision with the wall does not change the speed of block 2).
boat will initially touch the dock, as in Fig. 9-81; the boat can slide
through the water without significant resistance; both the car and
the boat can be approximated as uniform in their mass distribu-
tion. Determine what the width of the gap will be just as the car is
about to make the jump.
105 A 3.0 kg object moving at 8.0 m/s in the positive direc-
tion of an xaxis has a one-dimensional elastic collision with an ob-
ject of mass M, initially at rest. After the collision the object of
mass Mhas a velocity of 6.0 m/s in the positive direction of the
axis.What is mass M?
106 A 2140 kg railroad flatcar, which can move with negligible
friction, is motionless next to a platform. A 242 kg sumo wrestler
runs at 5.3 m/s along the platform (parallel to the track) and then
jumps onto the flatcar. What is the speed of the flatcar if he then
(a) stands on it, (b) runs at 5.3 m/s relative to it in his original direc-
tion, and (c) turns and runs at 5.3 m/s relative to the flatcar oppo-
site his original direction?
107 A 6100 kg rocket is set for vertical firing from the
ground. If the exhaust speed is 1200 m/s, how much gas must be
ejected each second if the thrust (a) is to equal the magnitude of
the gravitational force on the rocket and (b) is to give the rocket an
initial upward acceleration of 21 m/s2?
108 A 500.0 kg module is attached to a 400.0 kg shuttle craft,
which moves at 1000 m/s relative to the stationary main spaceship.
Then a small explosion sends the module backward with speed
100.0 m/s relative to the new speed of the shuttle craft. As meas-
ured by someone on the main spaceship, by what fraction did the
kinetic energy of the module and shuttle craft increase because of
the explosion?
109 (a) How far is the center of mass of the EarthMoon
system from the center of Earth? (Appendix C gives the masses of
Earth and the Moon and the distance between the two.) (b) What
percentage of Earth’s radius is that distance?
110 A 140 g ball with speed 7.8 m/s strikes a wall perpendicu-
larly and rebounds in the opposite direction with the same speed.
The collision lasts 3.80 ms. What are the magnitudes of the (a) im-
pulse and (b) average force on the wall from the ball during the
elastic collision?
111 A rocket sled with a mass of 2900 kg moves at 250 m/s
on a set of rails. At a certain point, a scoop on the sled dips into a
trough of water located between the tracks and scoops water into
an empty tank on the sled. By applying the principle of conserva-
tion of linear momentum, determine the speed of the sled after
920 kg of water has been scooped up.Ignore any retarding force on
the scoop.
112 A pellet gun fires ten 2.0 g pellets per second with a
speed of 500 m/s.The pellets are stopped by a rigid wall.What are
(a) the magnitude of the momentum of each pellet, (b) the ki-
netic energy of each pellet, and (c) the magnitude of the average
force on the wall from the stream of pellets? (d) If each pellet is
in contact with the wall for 0.60 ms, what is the magnitude of the
average force on the wall from each pellet during contact? (e)
Why is this average force so different from the average force cal-
culated in (c)?
113 A railroad car moves under a grain elevator at a constant
speed of 3.20 m/s. Grain drops into the car at the rate of 540 kg/min.
What is the magnitude of the force needed to keep the car moving
at constant speed if friction is negligible?
SSM
SSM
SSM
SSM
SSM
h
Figure 9-78 Problem 101.
Figure 9-79
Problem 102.
v2i
12
Figure 9-80 Problem 103.
104 The script for an action movie calls for a small race car (of
mass 1500 kg and length 3.0 m) to accelerate along a flattop boat
(of mass 4000 kg and length 14 m), from one end of the boat to the
other, where the car will then jump
the gap between the boat and a
somewhat lower dock. You are the
technical advisor for the movie. The
A45678SF
Dock Boat
Figure 9-81 Problem 104.
256 CHAPTER 9 CENTER OF MASS AND LINEAR MOMENTUM
v
VJ
m
M
Figure 9-84 Problem 123.
114 Figure 9-82 shows a uniform square plate of edge length
6d6.0 m from which a square piece of edge length 2dhas been
removed.What are (a) the xcoordinate and (b) the ycoordinate of
the center of mass of the remaining piece?
rates the body into two parts, each of 4.0 kg, and increases the total
kinetic energy by 16 J. The forward part continues to move in the
original direction of motion. What are the speeds of (a) the rear
part and (b) the forward part?
121 An electron undergoes a one-dimensional elastic collision
with an initially stationary hydrogen atom.What percentage of the
electron’s initial kinetic energy is transferred to kinetic energy of
the hydrogen atom? (The mass of the hydrogen atom is 1840 times
the mass of the electron.)
122 A man (weighing 915 N) stands on a long railroad flatcar
(weighing 2415 N) as it rolls at 18.2 m/s in the positive direction of
an xaxis, with negligible friction.Then the man runs along the flat-
car in the negative xdirection at 4.00 m/s relative to the flatcar.
What is the resulting increase in the speed of the flatcar?
123 An unmanned space probe (of mass mand speed vrelative to
the Sun) approaches the planet Jupiter (of mass Mand speed VJrel-
ative to the Sun) as shown in Fig. 9-84. The spacecraft rounds the
planet and departs in the opposite direction. What is its speed (in
kilometers per second), relative to the Sun, after this slingshot en-
counter, which can be analyzed as a collision? Assume v10.5 km/s
and VJ13.0 km/s (the orbital speed of Jupiter).The mass of Jupiter
is very much greater than the mass of the spacecraft (Mm).
115 At time t0, force N acts on an
initially stationary particle of mass 2.00 103kg and force
N acts on an initially stationary particle of
mass 4.00 103kg. From time t0 to t2.00 ms, what are the
(a) magnitude and (b) angle (relative to the positive direction of
the xaxis) of the displacement of the center of mass of the two-
particle system? (c) What is the kinetic energy of the center of
mass at t2.00 ms?
116 Two particles Pand Qare released from rest 1.0 m apart.Phas
a mass of 0.10 kg, and Qa mass of 0.30 kg. Pand Qattract each other
with a constant force of 1.0 102N. No external forces act on the
system. (a) What is the speed of the center of mass of Pand Qwhen
the separation is 0.50 m? (b) At what distance from P’s original posi-
tion do the particles collide?
117 A collision occurs between a 2.00 kg particle traveling with
velocity and a 4.00 kg particle
traveling with velocity . The colli-
sion connects the two particles. What then is their velocity in (a)
unit-vector notation and as a (b) magnitude and (c) angle?
118 In the two-sphere arrangement of Fig. 9-20, assume that
sphere 1 has a mass of 50 g and an initial height of h19.0 cm, and
that sphere 2 has a mass of 85 g.After sphere 1 is released and col-
lides elastically with sphere 2, what height is reached by (a) sphere
1 and (b) sphere 2? After the next (elastic) collision, what height is
reached by (c) sphere 1 and (d) sphere 2? (Hint: Do not use
rounded-off values.)
119 In Fig. 9-83, block 1 slides along
an xaxis on a frictionless floor with a
speed of 0.75 m/s. When it reaches sta-
tionary block 2, the two blocks undergo
an elastic collision. The following table
gives the mass and length of the (uni-
form) blocks and also the locations of their centers at time t0.
Where is the center of mass of the two-block system located (a) at
t0, (b) when the two blocks first touch, and (c) at t4.0 s?
Block Mass (kg) Length (cm) Center at t0
1 0.25 5.0 x1.50 m
2 0.50 6.0 x0
120 A body is traveling at 2.0 m/s along the positive direction of
an xaxis; no net force acts on the body. An internal explosion sepa-
(2.00 m/s)j
ˆ
v
:
2(6.00 m/s)i
ˆ
v
:
1(4.00 m/s)i
ˆ(5.00 m/s)j
ˆ
F
:
2(2.00i
ˆ4.00j
ˆ)
F
:
1(4.00i
ˆ5.00j
ˆ)
SSM
3d
3d
x
y
0
d
d
2d
2d
3d3d
Figure 9-82 Problem 114.
124 A 0.550 kg ball falls directly down onto concrete, hitting it
with a speed of 12.0 m/s and rebounding directly upward with a
speed of 3.00 m/s. Extend a yaxis upward. In unit-vector notation,
what are (a) the change in the ball’s momentum, (b) the impulse
on the ball, and (c) the impulse on the concrete?
125 An atomic nucleus at rest at the origin of an xy coordinate
system transforms into three particles. Particle 1, mass 16.7 1027
kg,moves away from the origin at velocity (6.00 106m/s) ; particle
2, mass 8.35 1027 kg, moves away at velocity (8.00 106m/s) .
(a) In unit-vector notation, what is the linear momentum of the
third particle, mass 11.7 1027 kg? (b) How much kinetic energy
appears in this transformation?
126 Particle 1 of mass 200 g and speed 3.00 m/s undergoes a one-
dimensional collision with stationary particle 2 of mass 400 g.What
is the magnitude of the impulse on particle 1 if the collision is (a)
elastic and (b) completely inelastic?
127 During a lunar mission, it is necessary to increase the speed
of a spacecraft by 2.2 m/s when it is moving at 400 m/s relative to
the Moon. The speed of the exhaust products from the rocket en-
gine is 1000 m/s relative to the spacecraft. What fraction of the
initial mass of the spacecraft must be burned and ejected to accom-
plish the speed increase?
128 A cue stick strikes a stationary pool ball, with an average
force of 32 N over a time of 14 ms. If the ball has mass 0.20 kg, what
speed does it have just after impact?
j
ˆ
i
ˆ
–1.50 m 0
x
1 2
Figure 9-83 Problem 119.
257
CHAPTER 10
Rotation
10-1ROTATIONAL VARIABLES
After reading this module, you should be able to . . .
10.01 Identify that if all parts of a body rotate around a fixed
axis locked together, the body is a rigid body. (This chapter
is about the motion of such bodies.)
10.02 Identify that the angular position of a rotating rigid body
is the angle that an internal reference line makes with a
fixed, external reference line.
10.03 Apply the relationship between angular displacement
and the initial and final angular positions.
10.04 Apply the relationship between average angular veloc-
ity, angular displacement, and the time interval for that dis-
placement.
10.05 Apply the relationship between average angular accel-
eration, change in angular velocity, and the time interval for
that change.
10.06 Identify that counterclockwise motion is in the positive
direction and clockwise motion is in the negative direction.
10.07 Given angular position as a function of time, calculate the
instantaneous angular velocity at any particular time and the
average angular velocity between any two particular times.
10.08 Given a graph of angular position versus time, deter-
mine the instantaneous angular velocity at a particular time
and the average angular velocity between any two particu-
lar times.
10.09 Identify instantaneous angular speed as the magnitude
of the instantaneous angular velocity.
10.10 Given angular velocity as a function of time, calculate
the instantaneous angular acceleration at any particular
time and the average angular acceleration between any
two particular times.
10.11 Given a graph of angular velocity versus time, deter-
mine the instantaneous angular acceleration at any partic-
ular time and the average angular acceleration between
any two particular times.
10.12 Calculate a body’s change in angular velocity by
integrating its angular acceleration function with respect
to time.
10.13 Calculate a body’s change in angular position by inte-
grating its angular velocity function with respect to time.
To describe the rotation of a rigid body about a fixed axis,
called the rotation axis, we assume a reference line is fixed in
the body, perpendicular to that axis and rotating with the
body. We measure the angular position uof this line
relative to a fixed direction. When uis measured in radians,
(radian measure),
where sis the arc length of a circular path of radius rand
angle u.
Radian measure is related to angle measure in revolutions
and degrees by
1 rev 3602prad.
A body that rotates about a rotation axis, changing its angu-
lar position from u1to u2, undergoes an angular displacement
uu2u1,
where uis positive for counterclockwise rotation and nega-
tive for clockwise rotation.
If a body rotates through an angular displacement uin a
time interval t, its average angular velocity vavg is
us
r
The (instantaneous) angular velocity vof the body is
Both vavg and vare vectors, with directions given by a
right-hand rule. They are positive for counterclockwise rota-
tion and negative for clockwise rotation. The magnitude of the
body’s angular velocity is the angular speed.
If the angular velocity of a body changes from v1to v2in a
time interval tt2t1, the average angular acceleration
aavg of the body is
The (instantaneous) angular acceleration aof the body is
Both aavg and aare vectors.
adv
dt .
aavg v2v1
t2t1
v
t.
vdu
dt .
vavg u
t.
Key Ideas
Learning Objectives
258 CHAPTER 10 ROTATION
What Is Physics?
As we have discussed, one focus of physics is motion. However, so far we
have examined only the motion of translation, in which an object moves along
a straight or curved line, as in Fig. 10-1a. We now turn to the motion of rotation,
in which an object turns about an axis, as in Fig. 10-1b.
You see rotation in nearly every machine, you use it every time you open a
beverage can with a pull tab, and you pay to experience it every time you go to an
amusement park. Rotation is the key to many fun activities, such as hitting a long
drive in golf (the ball needs to rotate in order for the air to keep it aloft longer)
and throwing a curveball in baseball (the ball needs to rotate in order for the air
to push it left or right). Rotation is also the key to more serious matters, such as
metal failure in aging airplanes.
We begin our discussion of rotation by defining the variables for the
motion, just as we did for translation in Chapter 2. As we shall see, the vari-
ables for rotation are analogous to those for one-dimensional motion and, as
in Chapter 2, an important special situation is where the acceleration (here the
rotational acceleration) is constant. We shall also see that Newton’s second
law can be written for rotational motion, but we must use a new quantity
called torque instead of just force. Work and the work–kinetic energy
theorem can also be applied to rotational motion, but we must use a new quan-
tity called rotational inertia instead of just mass. In short, much of what we
have discussed so far can be applied to rotational motion with, perhaps, a few
changes.
Caution: In spite of this repetition of physics ideas, many students find this
and the next chapter very challenging. Instructors have a variety of reasons as
to why, but two reasons stand out: (1) There are a lot of symbols (with Greek
Figure 10-1 Figure skater Sasha Cohen in motion of (a) pure translation in a fixed
direction and (b) pure rotation about a vertical axis.
(b)
(a)
Mike Segar/Reuters/Landov LLC
Elsa/Getty Images, Inc.
259
10-1 ROTATIONAL VARIABLES
letters) to sort out. (2) Although you are very familiar with linear motion (you
can get across the room and down the road just fine), you are probably very
unfamiliar with rotation (and that is one reason why you are willing to pay so
much for amusement park rides). If a homework problem looks like a foreign
language to you, see if translating it into the one-dimensional linear motion of
Chapter 2 helps. For example, if you are to find, say, an angular distance, tem-
porarily delete the word angular and see if you can work the problem with the
Chapter 2 notation and ideas.
Rotational Variables
We wish to examine the rotation of a rigid body about a fixed axis.A rigid body is
a body that can rotate with all its parts locked together and without any change in
its shape. A fixed axis means that the rotation occurs about an axis that does not
move.Thus, we shall not examine an object like the Sun, because the parts of the
Sun (a ball of gas) are not locked together. We also shall not examine an object
like a bowling ball rolling along a lane, because the ball rotates about a moving
axis (the ball’s motion is a mixture of rotation and translation).
Figure 10-2 shows a rigid body of arbitrary shape in rotation about a fixed
axis, called the axis of rotation or the rotation axis. In pure rotation (angular
motion), every point of the body moves in a circle whose center lies on the axis of
rotation, and every point moves through the same angle during a particular time
interval. In pure translation (linear motion), every point of the body moves in a
straight line, and every point moves through the same linear distance during a
particular time interval.
We deal nowone at a timewith the angular equivalents of the linear
quantities position, displacement, velocity, and acceleration.
Angular Position
Figure 10-2 shows a reference line, fixed in the body, perpendicular to the rotation
axis and rotating with the body. The angular position of this line is the angle of
the line relative to a fixed direction, which we take as the zero angular position.
In Fig.10-3, the angular position uis measured relative to the positive direction of
the xaxis. From geometry, we know that uis given by
(radian measure). (10-1)
Here sis the length of a circular arc that extends from the xaxis (the zero angular
position) to the reference line, and ris the radius of the circle.
us
r
Figure 10-2 A rigid body of arbitrary shape in pure rotation about the zaxis of a coordinate
system. The position of the reference line with respect to the rigid body is arbitrary, but it is
perpendicular to the rotation axis. It is fixed in the body and rotates with the body.
z
O
Reference line
Rotation
axis
x
y
Body This reference line is part of the body
and perpendicular to the rotation axis.
We use it to measure the rotation of the
body relative to a fixed direction.
Figure 10-3 The rotating rigid body of
Fig. 10-2 in cross section, viewed from
above. The plane of the cross section is
perpendicular to the rotation axis, which
now extends out of the page, toward you.
In this position of the body, the reference
line makes an angle uwith the xaxis.
x
y
Reference
line
θ
r
s
Rotation
axis
The body has rotated
counterclockwise
by angle . This is the
positive direction.
θ
This dot means that
the rotation axis is
out toward you.
An angle defined in this way is measured in radians (rad) rather than in
revolutions (rev) or degrees. The radian, being the ratio of two lengths, is a
pure number and thus has no dimension. Because the circumference of a circle of
radius ris 2pr, there are 2pradians in a complete circle:
(10-2)
and thus 1 rad 57.30.159 rev. (10-3)
We do not reset uto zero with each complete rotation of the reference line about
the rotation axis. If the reference line completes two revolutions from the zero
angular position, then the angular position uof the line is u4prad.
For pure translation along an xaxis, we can know all there is to know
about a moving body if we know x(t), its position as a function of time.
Similarly, for pure rotation, we can know all there is to know about a rotating
body if we know u(t), the angular position of the body’s reference line as a
function of time.
Angular Displacement
If the body of Fig. 10-3 rotates about the rotation axis as in Fig. 10-4, changing the
angular position of the reference line from u1to u2, the body undergoes an
angular displacement ugiven by
uu2u1. (10-4)
This definition of angular displacement holds not only for the rigid body as a
whole but also for every particle within that body.
Clocks Are Negative. If a body is in translational motion along an xaxis, its
displacement xis either positive or negative, depending on whether the body is
moving in the positive or negative direction of the axis. Similarly, the angular dis-
placement uof a rotating body is either positive or negative, according to the
following rule:
1 rev 360 2pr
r2p rad,
260 CHAPTER 10 ROTATION
An angular displacement in the counterclockwise direction is positive, and one in
the clockwise direction is negative.
Checkpoint 1
A disk can rotate about its central axis like a merry-go-round.Which of the following
pairs of values for its initial and final angular positions,respectively, give a negative
angular displacement: (a) 3 rad, 5 rad, (b) 3 rad, 7 rad, (c) 7 rad, 3 rad?
The phrase clocks are negative can help you remember this rule (they certainly
are negative when their alarms sound off early in the morning).
Angular Velocity
Suppose that our rotating body is at angular position u1at time t1and at
angular position u2at time t2as in Fig. 10-4. We define the average angular velocity
of the body in the time interval tfrom t1to t2to be
(10-5)
where uis the angular displacement during t(vis the lowercase omega).
vavg u2u1
t2t1
u
t,
261
10-1 ROTATIONAL VARIABLES
Figure 10-4 The reference line of the rigid body of Figs. 10-2 and 10-3 is at angular position
u1at time t1and at angular position u2at a later time t2. The quantity u(u2u1) is the
angular displacement that occurs during the interval t(t2t1).The body itself is not
shown.
x
y
Rotation axis
O
θ
1
θ
2
Δ
θ
At t2
At t1
Reference line
This change in the angle of the reference line
(which is part of the body) is equal to the angular
displacement of the body itself during this
time interval.
The (instantaneous) angular velocity v, with which we shall be most con-
cerned, is the limit of the ratio in Eq. 10-5 as tapproaches zero.Thus,
(10-6)
If we know u(t), we can find the angular velocity vby differentiation.
Equations 10-5 and 10-6 hold not only for the rotating rigid body as a whole
but also for every particle of that body because the particles are all locked
together. The unit of angular velocity is commonly the radian per second (rad/s)
or the revolution per second (rev/s). Another measure of angular velocity was
used during at least the first three decades of rock: Music was produced by vinyl
(phonograph) records that were played on turntables at or “45 rpm,
meaning at or 45 rev/min.
If a particle moves in translation along an xaxis, its linear velocity vis either
positive or negative, depending on its direction along the axis. Similarly, the angu-
lar velocity vof a rotating rigid body is either positive or negative, depending on
whether the body is rotating counterclockwise (positive) or clockwise (negative).
(“Clocks are negative” still works.) The magnitude of an angular velocity is called
the angular speed, which is also represented with v.
Angular Acceleration
If the angular velocity of a rotating body is not constant, then the body has an an-
gular acceleration. Let v2and v1be its angular velocities at times t2and t1,
respectively.The average angular acceleration of the rotating body in the interval
from t1to t2is defined as
(10-7)
in which vis the change in the angular velocity that occurs during the time
interval t. The (instantaneous) angular acceleration a, with which we shall be
most concerned, is the limit of this quantity as tapproaches zero.Thus,
(10-8)
As the name suggests, this is the angular acceleration of the body at a given in-
stant. Equations 10-7 and 10-8 also hold for every particle of that body. The unit of
angular acceleration is commonly the radian per second-squared (rad/s2) or the
revolution per second-squared (rev/s2).
alim
t:0
v
tdv
dt .
aavg v2v1
t2t1
v
t,
331
3 rev/min
331
3 rpm
vlim
t:0
u
tdu
dt .
262 CHAPTER 10 ROTATION
Calculations: To sketch the disk and its reference line at a
particular time, we need to determine ufor that time. To do
so, we substitute the time into Eq. 10-9. For t2.0 s, we get
This means that at t2.0 s the reference line on the disk
is rotated counterclockwise from the zero position by angle
1.2 rad 69(counterclockwise because uis positive). Sketch
1 in Fig.10-5bshows this position of the reference line.
Similarly, for t0, we find u1.00 rad 57,
which means that the reference line is rotated clockwise
from the zero angular position by 1.0 rad, or 57, as shown
in sketch 3. For t4.0 s, we find u0.60 rad 34
(sketch 5). Drawing sketches for when the curve crosses
the taxis is easy, because then u0 and the reference line
is momentarily aligned with the zero angular position
(sketches 2 and 4).
(b) At what time tmin does u(t) reach the minimum
value shown in Fig.10-5b? What is that minimum value?
1.2 rad 1.2 rad 360
2
rad 69.
u1.00 (0.600)(2.0) (0.250)(2.0)2
Sample Problem 10.01 Angular velocity derived from angular position
The disk in Fig. 10-5ais rotating about its central axis like a
merry-go-round. The angular position u(t) of a reference
line on the disk is given by
u1.00 0.600t0.250t2, (10-9)
with tin seconds, uin radians, and the zero angular position
as indicated in the figure. (If you like, you can translate all
this into Chapter 2 notation by momentarily dropping the
word “angular” from “angular position” and replacing the
symbol uwith the symbol x.What you then have is an equa-
tion that gives the position as a function of time, for the one-
dimensional motion of Chapter 2.)
(a) Graph the angular position of the disk versus time
from t3.0 s to t5.4 s. Sketch the disk and its angular
position reference line at t2.0 s, 0 s, and 4.0 s, and
when the curve crosses the taxis.
KEY IDEA
The angular position of the disk is the angular position
u(t) of its reference line, which is given by Eq. 10-9 as a function
of time t. So we graph Eq. 10-9; the result is shown in Fig. 10-5b.
A
Zero
angular
position
Reference
line
Rotation axis
(a)
(b)
2
0
–2 024 6
(rad)
(1) (2) (3) (4) (5)
t(s)
θ
–2
The angular position
of the disk is the angle
between these two lines.
Now, the disk is
at a zero angle.
θ
At t=−2 s, the disk
is at a positive
(counterclockwise)
angle. So, a positive
value is plotted.
This is a plot of the angle
of the disk versus time.
Now, it is at a
negative (clockwise)
angle. So, a negative
value is plotted.
θ
It has reversed
its rotation and
is again at a
zero angle.
Now, it is
back at a
positive
angle.
Figure 10-5 (a) A rotating disk. (b) A plot of the disk’s angular position u(t). Five sketches indicate the angular position of the refer-
ence line on the disk for five points on the curve. (c) A plot of the disk’s angular velocity v(t). Positive values of vcorrespond to
counterclockwise rotation, and negative values to clockwise rotation.
263
10-1 ROTATIONAL VARIABLES
t3.0 s to t6.0 s. Sketch the disk and indicate the direc-
tion of turning and the sign of vat t2.0 s,4.0 s,and tmin.
KEY IDEA
From Eq. 10-6, the angular velocity vis equal to du/dt as
given in Eq. 10-10. So, we have
v0.600 0.500t. (10-11)
The graph of this function v(t) is shown in Fig. 10-5c.
Because the function is linear, the plot is a straight line. The
slope is 0.500 rad/s2and the intercept with the vertical axis
(not shown) is 0.600 rad/s.
Calculations: To sketch the disk at t2.0 s, we substitute
that value into Eq. 10-11, obtaining
v1.6 rad/s. (Answer)
The minus sign here tells us that at t2.0 s, the disk is
turning clockwise (as indicated by the left-hand sketch in
Fig. 10-5c).
Substituting t4.0 s into Eq. 10-11 gives us
v1.4 rad/s. (Answer)
The implied plus sign tells us that now the disk is turning
counterclockwise (the right-hand sketch in Fig.10-5c).
For tmin, we already know that du/dt 0. So, we must
also have v0. That is, the disk momentarily stops when
the reference line reaches the minimum value of uin
Fig. 10-5b, as suggested by the center sketch in Fig. 10-5c. On
the graph of v versus tin Fig. 10-5c, this momentary stop is
the zero point where the plot changes from the negative
clockwise motion to the positive counterclockwise motion.
(d) Use the results in parts (a) through (c) to describe the
motion of the disk from t3.0 s to t6.0 s.
Description: When we first observe the disk at t3.0 s, it
has a positive angular position and is turning clockwise but
slowing. It stops at angular position u1.36 rad and then
begins to turn counterclockwise, with its angular position
eventually becoming positive again.
KEY IDEA
To find the extreme value (here the minimum) of a function,
we take the first derivative of the function and set the result
to zero.
Calculations: The first derivative of u(t) is
(10-10)
Setting this to zero and solving for tgive us the time at
which u(t) is minimum:
tmin 1.20 s. (Answer)
To get the minimum value of u, we next substitute tmin into
Eq. 10-9, finding
u1.36 rad 77.9. (Answer)
This minimum of u(t) (the bottom of the curve in Fig. 10-5b)
corresponds to the maximum clockwise rotation of the disk
from the zero angular position, somewhat more than is
shown in sketch 3.
(c) Graph the angular velocity vof the disk versus time from
du
dt 0.600 0.500t.
(c)
2
0
–2 –2 0 2 4 6
(rad/s)
ω
t(s)
negative
ω
zero
ω
positive
ω
This is a plot of the angular
velocity of the disk versus time.
The angular velocity is
initially negative and slowing,
then momentarily zero during
reversal, and then positive and
increasing.
Additional examples, video, and
practice available at WileyPLUS
264 CHAPTER 10 ROTATION
Are Angular Quantities Vectors?
We can describe the position, velocity, and acceleration of a single particle by
means of vectors. If the particle is confined to a straight line, however, we do not
really need vector notation. Such a particle has only two directions available to it,
and we can indicate these directions with plus and minus signs.
In the same way, a rigid body rotating about a fixed axis can rotate only
clockwise or counterclockwise as seen along the axis, and again we can select
between the two directions by means of plus and minus signs.The question arises:
“Can we treat the angular displacement, velocity, and acceleration of a rotating
body as vectors?” The answer is a qualified “yes” (see the caution below, in con-
nection with angular displacements).
Angular Velocities. Consider the angular velocity. Figure 10-6ashows a
vinyl record rotating on a turntable. The record has a constant angular speed
in the clockwise direction. We can represent its angular ve-
locity as a vector pointing along the axis of rotation, as in Fig. 10-6b. Here’s
how: We choose the length of this vector according to some convenient scale,
for example, with 1 cm corresponding to 10 rev/min.Then we establish a direc-
tion for the vector by using a right-hand rule, as Fig. 10-6cshows: Curl your
right hand about the rotating record, your fingers pointing in the direction of
rotation. Your extended thumb will then point in the direction of the angular
velocity vector. If the record were to rotate in the opposite sense, the right-
v
:
v
:
v
(331
3 rev/min)
To evaluate the constant of integration C, we note that v
5 rad/s at t0. Substituting these values in our expression
for vyields
,
so C5 rad/s.Then
. (Answer)
(b) Obtain an expression for the angular position u(t) of the
top.
KEY IDEA
By definition, v(t) is the derivative of u(t) with respect to
time. Therefore, we can find u(t) by integrating v(t) with
respect to time.
Calculations: Since Eq. 10-6 tells us that
duv dt,
we can write
(Answer)
where Chas been evaluated by noting that u2 rad at t0.
1
4t52
3t35t2,
1
4t52
3t35tC
uv
dt (5
4t42t25) dt
v5
4t42t25
5 rad/s 0 0 C
Sample Problem 10.02 Angular velocity derived from angular acceleration
A child’s top is spun with angular acceleration
,
with tin seconds and ain radians per second-squared. At
t0, the top has angular velocity 5 rad/s, and a reference
line on it is at angular position u2 rad.
(a) Obtain an expression for the angular velocity v(t) of the
top.That is, find an expression that explicitly indicates how the
angular velocity depends on time. (We can tell that there is
such a dependence because the top is undergoing an angular
acceleration,which means that its angular velocity is changing.)
KEY IDEA
By definition, a(t) is the derivative of v(t) with respect to time.
Thus, we can find v(t) by integrating a(t) with respect to time.
Calculations: Equation 10-8 tells us
,
so .
From this we find
.v(5t34t)dt 5
4t44
2t2C
dva
dt
dvadt
a5t34t
Additional examples, video, and practice available at WileyPLUS
265
10-1 ROTATIONAL VARIABLES
hand rule would tell you that the angular velocity vector then points in the op-
posite direction.
It is not easy to get used to representing angular quantities as vectors. We in-
stinctively expect that something should be moving along the direction of a vec-
tor. That is not the case here. Instead, something (the rigid body) is rotating
around the direction of the vector. In the world of pure rotation, a vector defines
an axis of rotation, not a direction in which something moves. Nonetheless, the
vector also defines the motion. Furthermore, it obeys all the rules for vector
manipulation discussed in Chapter 3. The angular acceleration is another
vector,and it too obeys those rules.
In this chapter we consider only rotations that are about a fixed axis. For such
situations, we need not consider vectorswe can represent angular velocity with
vand angular acceleration with a, and we can indicate direction with an implied
plus sign for counterclockwise or an explicit minus sign for clockwise.
Angular Displacements.Now for the caution: Angular displacements
(unless they are very small) cannot be treated as vectors. Why not? We can cer-
tainly give them both magnitude and direction, as we did for the angular veloc-
ity vector in Fig. 10-6. However, to be represented as a vector, a quantity must
also obey the rules of vector addition, one of which says that if you add two
vectors, the order in which you add them does not matter. Angular displace-
ments fail this test.
Figure 10-7 gives an example. An initially horizontal book is given two
90angular displacements, first in the order of Fig. 10-7aand then in the order
of Fig. 10-7b.Although the two angular displacements are identical, their order
is not, and the book ends up with different orientations. Here’s another exam-
ple. Hold your right arm downward, palm toward your thigh. Keeping your
wrist rigid, (1) lift the arm forward until it is horizontal, (2) move it horizon-
tally until it points toward the right, and (3) then bring it down to your side.
Your palm faces forward. If you start over, but reverse the steps, which way
does your palm end up facing? From either example, we must conclude that
the addition of two angular displacements depends on their order and they
cannot be vectors.
a
:
Figure 10-6 (a) A record rotating about a vertical axis that coincides with the axis of the
spindle. (b) The angular velocity of the rotating record can be represented by the vector
, lying along the axis and pointing down, as shown. (c) We establish the direction of the
angular velocity vector as downward by using a right-hand rule. When the fingers of the
right hand curl around the record and point the way it is moving, the extended thumb
points in the direction of .v
:
v
:
zz z
(a) (b) (c)
Axis Axis Axis
ω
Spindle
ω
This right-hand rule
establishes the
direction of the
angular velocity
vector.
Figure 10-7 (a) From its initial position, at
the top, the book is given two successive
90rotations, first about the (horizontal)
xaxis and then about the (vertical) yaxis.
(b) The book is given the same rotations,
but in the reverse order.
PHYSICS
PHYSICS
PHYSICS
PHYSICS
PHYSICS
PHYSICS
PHYSICS
(a) (b)
PHYSICS
y
x
z
y
x
z
y
x
zz
y
x
z
y
x
y
x
zThe order of the
rotations makes
a big difference
in the result.
266 CHAPTER 10 ROTATION
Rotation with Constant Angular Acceleration
In pure translation, motion with a constant linear acceleration (for example, that
of a falling body) is an important special case. In Table 2-1, we displayed a series
of equations that hold for such motion.
In pure rotation, the case of constant angular acceleration is also important,
and a parallel set of equations holds for this case also. We shall not derive them
here, but simply write them from the corresponding linear equations, substituting
equivalent angular quantities for the linear ones.This is done in Table 10-1, which
lists both sets of equations (Eqs. 2-11 and 2-15 to 2-18; 10-12 to 10-16).
Recall that Eqs. 2-11 and 2-15 are basic equations for constant linear
accelerationthe other equations in the Linear list can be derived from them.
Similarly, Eqs. 10-12 and 10-13 are the basic equations for constant angular
acceleration, and the other equations in the Angular list can be derived from
them.To solve a simple problem involving constant angular acceleration, you can
usually use an equation from the Angular list (if you have the list). Choose
an equation for which the only unknown variable will be the variable requested
in the problem.A better plan is to remember only Eqs. 10-12 and 10-13, and then
solve them as simultaneous equations whenever needed.
10-2 ROTATION WITH CONSTANT ANGULAR ACCELERATION
After reading this module, you should be able to . . .
10.14 For constant angular acceleration, apply the relation-
ships between angular position, angular displacement,
Key Idea
Constant angular acceleration (aconstant) is an important special case of rotational motion. The appropriate kinematic
equations are
vv0at,
uu0vt1
2at2.
uu01
2(v0v)t,
v2v0
22a(uu0),
uu0v0t1
2at2,
Learning Objective
angular velocity, angular acceleration, and elapsed time
(Table 10-1).
Table 10-1 Equations of Motion for Constant Linear Acceleration and for Constant Angular Acceleration
Equation Linear Missing Angular Equation
Number Equation Variable Equation Number
(2-11) vv0at x x0uu0vv0at(10-12)
(2-15) vv(10-13)
(2-16) tt (10-14)
(2-17) aa(10-15)
(2-18) v0v0(10-16)uu0vt1
2at2
xx0vt 1
2at2
uu01
2(v0v)tx x01
2(v0v)t
v2v0
22a(uu0)v2v0
22a(xx0)
uu0v0t1
2at2
xx0v0t1
2at2
Checkpoint 2
In four situations, a rotating body has angular position u(t) given by (a) u3t4,
(b) u5t34t26, (c) u2/t24/t,and (d) u5t23.To which situations do
the angular equations of Table 10-1 apply?
267
10-2 ROTATION WITH CONSTANT ANGULAR ACCELERATION
(We converted 5.0 rev to 10prad to keep the units consis-
tent.) Solving this quadratic equation for t,we find
t32 s. (Answer)
Now notice something a bit strange. We first see the wheel
when it is rotating in the negative direction and through the
u0 orientation.Yet, we just found out that 32 s later it is at
the positive orientation of u5.0 rev. What happened in
that time interval so that it could be at a positive orientation?
(b) Describe the grindstone’s rotation between t0 and
t32 s.
Description: The wheel is initially rotating in the negative
(clockwise) direction with angular velocity v04.6 rad/s,
but its angular acceleration ais positive.This initial opposi-
tion of the signs of angular velocity and angular accelera-
tion means that the wheel slows in its rotation in the nega-
tive direction, stops, and then reverses to rotate in the
positive direction. After the reference line comes back
through its initial orientation of u0, the wheel turns an
additional 5.0 rev by time t32 s.
(c) At what time tdoes the grindstone momentarily stop?
Calculation: We again go to the table of equations for con-
stant angular acceleration, and again we need an equation
that contains only the desired unknown variable t. However,
now the equation must also contain the variable v, so that we
can set it to 0 and then solve for the corresponding time t. We
choose Eq. 10-12,which yields
(Answer)tvv0
a0(4.6 rad/s)
0.35 rad/s213 s.
Sample Problem 10.03 Constant angular acceleration, grindstone
A grindstone (Fig. 10-8) rotates at constant angular acceler-
ation a0.35 rad/s2. At time t0, it has an angular velocity
of v04.6 rad/s and a reference line on it is horizontal, at
the angular position u00.
(a) At what time after t0 is the reference line at the
angular position u5.0 rev?
KEY IDEA
The angular acceleration is constant, so we can use the rota-
tion equations of Table 10-1.We choose Eq.10-13,
,
because the only unknown variable it contains is the desired
time t.
Calculations: Substituting known values and setting u00
and u5.0 rev 10prad give us
.10p rad (4.6 rad/s)t1
2 (0.35 rad/s2)t2
uu0v0t1
2at2
Figure 10-8 A grindstone.At t0 the reference line (which we
imagine to be marked on the stone) is horizontal.
Axis
Reference
line
Zero angular
position
We measure rotation by using
this reference line.
Clockwise = negative
Counterclockwise = positive
rad/s, the angular displacement is uu020.0 rev, and the
angular velocity at the end of that displacement is v2.00
rad/s. In addition to the angular acceleration athat we want,
both basic equations also contain time t, which we do not
necessarily want.
To eliminate the unknown t, we use Eq. 10-12 to write
which we then substitute into Eq. 10-13 to write
Solving for a, substituting known data, and converting
20 rev to 125.7 rad, we find
(Answer)0.0301 rad/s2.
av2v0
2
2(uu0)(2.00 rad/s)2(3.40 rad/s)2
2(125.7 rad)
uu0v0
vv0
a
1
2a
vv0
a
2.
tvv0
a,
Sample Problem 10.04 Constant angular acceleration, riding a Rotor
While you are operating a Rotor (a large, vertical, rotating
cylinder found in amusement parks), you spot a passenger in
acute distress and decrease the angular velocity of the cylin-
der from 3.40 rad/s to 2.00 rad/s in 20.0 rev, at constant angu-
lar acceleration. (The passenger is obviously more of a “trans-
lation person” than a “rotation person.”)
(a) What is the constant angular acceleration during this
decrease in angular speed?
KEY IDEA
Because the cylinder’s angular acceleration is constant, we
can relate it to the angular velocity and angular displacement
via the basic equations for constant angular acceleration
(Eqs. 10-12 and 10-13).
Calculations: Let’s first do a quick check to see if we can solve
the basic equations. The initial angular velocity is v03.40
268 CHAPTER 10 ROTATION
Relating the Linear and Angular Variables
In Module 4-5,we discussed uniform circular motion,in which a particle travels at con-
stant linear speed valong a circle and around an axis of rotation. When a rigid body,
such as a merry-go-round,rotates around an axis,each particle in the body moves in its
own circle around that axis. Since the body is rigid, all the particles make one revolu-
tion in the same amount of time;that is,they all have the same angular speed v.
However, the farther a particle is from the axis, the greater the circumference
of its circle is, and so the faster its linear speed vmust be.You can notice this on a
merry-go-round. You turn with the same angular speed vregardless of your dis-
tance from the center, but your linear speed vincreases noticeably if you move to
the outside edge of the merry-go-round.
We often need to relate the linear variables s,v, and afor a particular point in
a rotating body to the angular variables u,v, and afor that body. The two sets of
variables are related by r, the perpendicular distance of the point from the
rotation axis. This perpendicular distance is the distance between the point and
the rotation axis, measured along a perpendicular to the axis. It is also the radius r
of the circle traveled by the point around the axis of rotation.
(b) How much time did the speed decrease take?
Calculation: Now that we know a, we can use Eq. 10-12 to
solve for t:(Answer)46.5 s.
tvv0
a2.00 rad/s 3.40 rad/s
0.0301 rad/s2
10-3 RELATING THE LINEAR AND ANGULAR VARIABLES
After reading this module, you should be able to . . .
10.15 For a rigid body rotating about a fixed axis, relate the angular
variables of the body (angular position, angular velocity, and an-
gular acceleration) and the linear variables of a particle on the
body (position, velocity, and acceleration) at any given radius.
10.16 Distinguish between tangential acceleration and radial
acceleration, and draw a vector for each in a sketch of a
particle on a body rotating about an axis, for both an in-
crease in angular speed and a decrease.
A point in a rigid rotating body, at a perpendicular distance
rfrom the rotation axis, moves in a circle with radius r. If the
body rotates through an angle u, the point moves along an
arc with length sgiven by
sur(radian measure),
where uis in radians.
The linear velocity of the point is tangent to the circle; the
point’s linear speed vis given by
vvr(radian measure),
where vis the angular speed (in radians per second) of the body,
and thus also the point.
v
:
The linear acceleration of the point has both tangential
and radial components. The tangential component is
atar(radian measure),
where ais the magnitude of the angular acceleration (in radi-
ans per second-squared) of the body. The radial component
of is
(radian measure).
If the point moves in uniform circular motion, the period Tof
the motion for the point and the body is
(radian measure).T2pr
v2p
v
arv2
rv2r
a
:
a
:
Learning Objectives
Key Ideas
Additional examples, video, and practice available at WileyPLUS
269
10-3 RELATING THE LINEAR AND ANGULAR VARIABLES
The Position
If a reference line on a rigid body rotates through an angle u, a point within the
body at a position rfrom the rotation axis moves a distance salong a circular arc,
where sis given by Eq. 10-1:
sur(radian measure). (10-17)
This is the first of our linearangular relations. Caution: The angle uhere must be
measured in radians because Eq. 10-17 is itself the definition of angular measure
in radians.
The Speed
Differentiating Eq. 10-17 with respect to timewith rheld constantleads to
However, ds/dt is the linear speed (the magnitude of the linear velocity) of the
point in question, and du/dt is the angular speed vof the rotating body. So
vvr(radian measure). (10-18)
Caution: The angular speed vmust be expressed in radian measure.
Equation 10-18 tells us that since all points within the rigid body have the
same angular speed v, points with greater radius rhave greater linear speed v.
Figure 10-9areminds us that the linear velocity is always tangent to the circular
path of the point in question.
If the angular speed vof the rigid body is constant, then Eq. 10-18 tells
us that the linear speed vof any point within it is also constant. Thus, each point
within the body undergoes uniform circular motion. The period of revolution T
for the motion of each point and for the rigid body itself is given by Eq. 4-35:
. (10-19)
This equation tells us that the time for one revolution is the distance 2prtraveled
in one revolution divided by the speed at which that distance is traveled.
Substituting for vfrom Eq. 10-18 and canceling r, we find also that
(radian measure). (10-20)
This equivalent equation says that the time for one revolution is the angular dis-
tance 2prad traveled in one revolution divided by the angular speed (or rate) at
which that angle is traveled.
The Acceleration
Differentiating Eq. 10-18 with respect to timeagain with rheld constant
leads to
(10-21)
Here we run up against a complication. In Eq. 10-21, dv/dt represents only the
part of the linear acceleration that is responsible for changes in the magnitude v
of the linear velocity . Like , that part of the linear acceleration is tangent to
the path of the point in question.We call it the tangential component atof the lin-
ear acceleration of the point, and we write
atar(radian measure), (10-22)
v
:
v
:
dv
dt dv
dt r.
T2p
v
T2pr
v
ds
dt du
dt r.
Figure 10-9 The rotating rigid body of Fig. 10-2,
shown in cross section viewed from above.
Every point of the body (such as P) moves
in a circle around the rotation axis. (a)The
linear velocity of every point is tangent to
the circle in which the point moves. (b) The
linear acceleration of the point has (in
general) two components: tangential atand
radial ar.
a
:
v
:
x
y
r
Rotation
axis
P
Circle
traveled by P
(a)
v
The velocity vector is
always tangent to this
circle around the
rotation axis.
x
y
ar
P
(b)
at
Rotation
axis
The acceleration always
has a radial (centripetal)
component and may have
a tangential component.
270 CHAPTER 10 ROTATION
where adv/dt. Caution: The angular acceleration ain Eq. 10-22 must be
expressed in radian measure.
In addition, as Eq. 4-34 tells us, a particle (or point) moving in a circular path
has a radial component of linear acceleration, arv2/r(directed radially inward),
that is responsible for changes in the direction of the linear velocity . By substi-
tuting for vfrom Eq. 10-18, we can write this component as
(radian measure). (10-23)
Thus, as Fig. 10-9bshows, the linear acceleration of a point on a rotating rigid
body has, in general, two components. The radially inward component ar(given
by Eq. 10-23) is present whenever the angular velocity of the body is not zero.
The tangential component at(given by Eq. 10-22) is present whenever the angu-
lar acceleration is not zero.
arv2
rv2r
v
:
Checkpoint 3
A cockroach rides the rim of a rotating merry-go-round. If the angular speed of this
system (merry-go-round cockroach) is constant,does the cockroach have (a) radial
acceleration and (b) tangential acceleration? If vis decreasing,does the cockroach
have (c) radial acceleration and (d) tangential acceleration?
and radial accelerations are the (perpendicular) compo-
nents of the (full) acceleration .
Calculations: Let’s go through the steps. We first find the
angular velocity by taking the time derivative of the given
angular position function and then substituting the given
time of t2.20 s:
v(ct3)3ct2(10-25)
3(6.39 102rad/s3)(2.20 s)2
0.928 rad/s. (Answer)
From Eq. 10-18, the linear speed just then is
vvr3ct2r(10-26)
3(6.39 102rad/s3)(2.20 s)2(33.1 m)
30.7 m/s. (Answer)
du
dt d
dt
a
:
Sample Problem 10.05 Designing The Giant Ring, a large-scale amusement park ride
We are given the job of designing a large horizontal ring
that will rotate around a vertical axis and that will have a ra-
dius of r33.1 m (matching that of Beijing’s The Great
Observation Wheel, the largest Ferris wheel in the world).
Passengers will enter through a door in the outer wall of the
ring and then stand next to that wall (Fig.10-10a).We decide
that for the time interval t0 to t2.30 s, the angular posi-
tion u(t) of a reference line on the ring will be given by
uct3, (10-24)
with c6.39 102rad/s3. After t2.30 s, the angular
speed will be held constant until the end of the ride. Once
the ring begins to rotate, the floor of the ring will drop away
from the riders but the riders will not fall—indeed, they feel
as though they are pinned to the wall. For the time t2.20 s,
let’s determine a rider’s angular speed v, linear speed v,an-
gular acceleration a, tangential acceleration at, radial accel-
eration ar, and acceleration .
KEY IDEAS
(1) The angular speed vis given by Eq. 10-6 (vdu/dt).
(2) The linear speed v(along the circular path) is related to
the angular speed (around the rotation axis) by Eq. 10-18
(vvr). (3) The angular acceleration ais given by Eq. 10-8
(adv/dt). (4) The tangential acceleration at(along the cir-
cular path) is related to the angular acceleration (around
the rotation axis) by Eq. 10-22 (atar). (5) The radial accel-
eration aris given Eq. 10-23 (arv2r). (6) The tangential
a
:
u
a
ar
at
(b)(a)
Figure 10-10 (a) Overhead view of
a passenger ready to ride The
Giant Ring. (b) The radial and
tangential acceleration compo-
nents of the (full) acceleration.
271
10-4 KINETIC ENERGY OF ROTATION
Additional examples, video, and practice available at WileyPLUS
The radial and tangential accelerations are perpendicu-
lar to each other and form the components of the rider’s
acceleration (Fig. 10-10b). The magnitude of is given by
a(10-29)
39.9 m/s2, (Answer)
or 4.1g(which is really exciting!). All these values are
acceptable.
To find the orientation of , we can calculate the angle u
shown in Fig.10-10b:
tan u
However, instead of substituting our numerical results, let’s
use the algebraic results from Eqs. 10-27 and 10-28:
utan1. (10-30)
The big advantage of solving for the angle algebraically is that
we can then see that the angle (1) does not depend on the
ring’s radius and (2) decreases as tgoes from 0 to 2.20 s. That
is, the acceleration vector swings toward being radially in-
ward because the radial acceleration (which depends on t4)
quickly dominates over the tangential acceleration (which
depends on only t).At our given time t2.20 s,we have
u . (Answer)tan12
3(6.39 102 rad/s3)(2.20 s)344.4
a
:
6ctr
9c2t4r
tan1
2
3ct3
at
ar
.
a
:
2(28.49 m/s2)2(27.91 m/s2)2
2a2
ra2
t
a
:
a
:
Although this is fast (111 km/h or 68.7 mi/h), such speeds are
common in amusement parks and not alarming because (as
mentioned in Chapter 2) your body reacts to accelerations but
not to velocities. (It is an accelerometer, not a speedometer.)
From Eq. 10-26 we see that the linear speed is increasing as the
square of the time (but this increase will cut off at t2.30 s).
Next, let’s tackle the angular acceleration by taking the
time derivative of Eq. 10-25:
a(3ct2)6ct
6(6.39 102rad/s3)(2.20 s) 0.843 rad/s2. (Answer)
The tangential acceleration then follows from Eq. 10-22:
atar6ctr (10-27)
6(6.39 102rad/s3)(2.20 s)(33.1 m)
27.91 m/s227.9 m/s2, (Answer)
or 2.8g(which is reasonable and a bit exciting). Equation
10-27 tells us that the tangential acceleration is increasing
with time (but it will cut off at t2.30 s). From Eq. 10-23,
we write the radial acceleration as
arv2r.
Substituting from Eq. 10-25 leads us to
ar(3ct2)2r9c2t4r(10-28)
9(6.39 102rad/s3)2(2.20 s)4(33.1 m)
28.49 m/s228.5 m/s2, (Answer)
or 2.9g(which is also reasonable and a bit exciting).
dv
dt d
dt
10-4 KINETIC ENERGY OF ROTATION
After reading this module, you should be able to . . .
10.17 Find the rotational inertia of a particle about a point.
10.18 Find the total rotational inertia of many particles moving
around the same fixed axis.
10.19 Calculate the rotational kinetic energy of a
body in terms of its rotational inertia and its angular
speed.
The kinetic energy Kof a rigid body rotating about a fixed
axis is given by
(radian measure),K1
2Iv2
in which Iis the rotational inertia of the body, defined as
for a system of discrete particles.
I
miri
2
Learning Objectives
Key Idea
Kinetic Energy of Rotation
The rapidly rotating blade of a table saw certainly has kinetic energy due to that
rotation. How can we express the energy? We cannot apply the familiar formula
to the saw as a whole because that would give us the kinetic energy
only of the saw’s center of mass, which is zero.
K1
2mv2
272 CHAPTER 10 ROTATION
Figure 10-11 A long rod is much easier to
rotate about (a) its central (longitudinal)
axis than about (b) an axis through its
center and perpendicular to its length. The
reason for the difference is that the mass
is distributed closer to the rotation axis in
(a) than in (b).
Rotation
axis
(a)
(b)
Rod is easy to rotate
this way.
Harder this way.
Instead, we shall treat the table saw (and any other rotating rigid body) as a
collection of particles with different speeds. We can then add up the kinetic
energies of all the particles to find the kinetic energy of the body as a whole.
In this way we obtain, for the kinetic energy of a rotating body,
(10-31)
in which miis the mass of the ith particle and viis its speed.The sum is taken over
all the particles in the body.
The problem with Eq. 10-31 is that viis not the same for all particles.We solve
this problem by substituting for vfrom Eq. 10-18 (vvr), so that we have
(10-32)
in which vis the same for all particles.
The quantity in parentheses on the right side of Eq. 10-32 tells us how
the mass of the rotating body is distributed about its axis of rotation. We call
that quantity the rotational inertia (or moment of inertia)Iof the body with
respect to the axis of rotation. It is a constant for a particular rigid body and
a particular rotation axis. (Caution: That axis must always be specified if the
value of Iis to be meaningful.)
We may now write
(rotational inertia) (10-33)
and substitute into Eq. 10-32, obtaining
(radian measure) (10-34)
as the expression we seek. Because we have used the relation vvrin deriving
Eq. 10-34, vmust be expressed in radian measure. The SI unit for Iis the
kilogramsquare meter (kgm2).
The Plan. If we have a few particles and a specified rotation axis, we find mr2
for each particle and then add the results as in Eq. 10-33 to get the total rotational in-
ertia I. If we want the total rotational kinetic energy, we can then substitute that I
into Eq. 10-34.That is the plan for a few particles, but suppose we have a huge num-
ber of particles such as in a rod. In the next module we shall see how to handle such
continuous bodies and do the calculation in only a few minutes.
Equation 10-34, which gives the kinetic energy of a rigid body in pure rotation,
is the angular equivalent of the formula , which gives the kinetic energyK1
2Mvcom
2
K1
2I
2
I
miri
2
K
1
2mi(vri)21
2
miri
2
v2,
1
2mivi
2,
K1
2m1v2
11
2m2v2
21
2m3v2
3 
of a rigid body in pure translation. In both formulas there is a factor of . Where
mass Mappears in one equation, I(which involves both mass and its distribution)
appears in the other. Finally, each equation contains as a factor the square of a
speedtranslational or rotational as appropriate. The kinetic energies of transla-
tion and of rotation are not different kinds of energy. They are both kinetic energy,
expressed in ways that are appropriate to the motion at hand.
We noted previously that the rotational inertia of a rotating body involves
not only its mass but also how that mass is distributed. Here is an example that
you can literally feel. Rotate a long, fairly heavy rod (a pole, a length of lumber,
or something similar), first around its central (longitudinal) axis (Fig. 10-11a)
and then around an axis perpendicular to the rod and through the center
(Fig. 10-11b). Both rotations involve the very same mass, but the first rotation is
much easier than the second. The reason is that the mass is distributed much
closer to the rotation axis in the first rotation. As a result, the rotational inertia
of the rod is much smaller in Fig. 10-11athan in Fig. 10-11b. In general, smaller
rotational inertia means easier rotation.
1
2
273
10-5 CALCULATING THE ROTATIONAL INERTIA
Checkpoint 4
The figure shows three small spheres that rotate
about a vertical axis.The perpendicular distance
between the axis and the center of each sphere is
given. Rank the three spheres according to their
rotational inertia about that axis,greatest first.
Rotation
axis
4 kg
3 m
2 m
1 m
9 kg
36 kg
10-5 CALCULATING THE ROTATIONAL INERTIA
After reading this module, you should be able to . . .
10.20 Determine the rotational inertia of a body if it is given in
Table 10-2.
10.21 Calculate the rotational inertia of a body by integration
over the mass elements of the body.
10.22 Apply the parallel-axis theorem for a rotation axis that is
displaced from a parallel axis through the center of mass of
a body.
Iis the rotational inertia of the body, defined as
for a system of discrete particles and defined as
for a body with continuously distributed mass. The rand riin
these expressions represent the perpendicular distance from
the axis of rotation to each mass element in the body, and the
integration is carried out over the entire body so as to include
every mass element.
Ir2dm
I
miri
2
The parallel-axis theorem relates the rotational inertia Iof a
body about any axis to that of the same body about a parallel
axis through the center of mass:
IIcom Mh2.
Here his the perpendicular distance between the two axes,
and Icom is the rotational inertia of the body about the axis
through the com. We can describe has being the distance
the actual rotation axis has been shifted from the rotation axis
through the com.
Learning Objectives
Key Ideas
Calculating the Rotational Inertia
If a rigid body consists of a few particles, we can calculate its rotational inertia
about a given rotation axis with Eq. 10-33 ; that is, we can find the
product mr2for each particle and then sum the products. (Recall that ris the per-
pendicular distance a particle is from the given rotation axis.)
If a rigid body consists of a great many adjacent particles (it is continuous, like
a Frisbee), using Eq. 10-33 would require a computer.Thus, instead, we replace the
sum in Eq. 10-33 with an integral and define the rotational inertia of the body as
(rotational inertia, continuous body). (10-35)
Table 10-2 gives the results of such integration for nine common body shapes and
the indicated axes of rotation.
Parallel-Axis Theorem
Suppose we want to find the rotational inertia Iof a body of mass Mabout a
given axis. In principle, we can always find Iwith the integration of Eq. 10-35.
However, there is a neat shortcut if we happen to already know the rotational in-
ertia Icom of the body about a parallel axis that extends through the body’s center
of mass. Let hbe the perpendicular distance between the given axis and the axis
Ir2dm
(Imiri
2)
274 CHAPTER 10 ROTATION
Table 10-2 Some Rotational Inertias
Axis
Hoop about
central axis
Axis
Annular cylinder
(or ring) about
central axis
R
I=MR2(b)(a)I=M(R1
2+R2
2)
R2
R1
Thin rod about
axis through center
perpendicular to
length
(e)
I=ML2
L
Axis
Axis
Axis
Hoop about any
diameter
Slab about
perpendicular
axis through
center
(i)(h)
I=MR2I= M(a2+b2)
R
b
a
Axis
Solid cylinder
(or disk) about
central axis
(c)
I=MR2
R
L
Axis
Solid cylinder
(or disk) about
central diameter
(d)
I=MR2+ML2
R
L
Axis
Thin
spherical shell
about any
diameter
(g)
I=MR2
2R
Solid sphere
about any
diameter
(f)
I=MR2
2R
Axis
1
__
21
__
2
2
__
5
1
__
4
2
__
31
__
2
1
__
12
1
__
12
1
__
12
Figure 10-12 A rigid body in cross section,
with its center of mass at O.The parallel-
axis theorem (Eq. 10-36) relates the
rotational inertia of the body about an axis
through Oto that about a parallel axis
through a point such as P, a distance h
from the body’s center of mass.
dm
r
P
h
a
b
x a
y b
com
O
Rotation axis
through
center of mass
Rotation axis
through P
y
x
We need to relate the rotational inertia
around the axis at P to that around the
axis at the com.
through the center of mass (remember these two axes must be parallel).Then the
rotational inertia Iabout the given axis is
IIcom Mh2(parallel-axis theorem). (10-36)
Think of the distance has being the distance we have shifted the rotation axis
from being through the com.This equation is known as the parallel-axis theorem.
We shall now prove it.
Proof of the Parallel-Axis Theorem
Let Obe the center of mass of the arbitrarily shaped body shown in cross section
in Fig. 10-12. Place the origin of the coordinates at O. Consider an axis through O
perpendicular to the plane of the figure, and another axis through point Pparal-
lel to the first axis. Let the xand ycoordinates of Pbe aand b.
Let dm be a mass element with the general coordinates xand y. The rota-
tional inertia of the body about the axis through Pis then, from Eq. 10-35,
which we can rearrange as
(10-37)
From the definition of the center of mass (Eq. 9-9), the middle two integrals of
Eq. 10-37 give the coordinates of the center of mass (multiplied by a constant)
I (x2y2)dm 2axdm2bydm(a2b2)dm.
Ir2dm [(xa)2(yb)2]dm,
275
10-5 CALCULATING THE ROTATIONAL INERTIA
and thus must each be zero. Because x2y2is equal to R2, where Ris the dis-
tance from Oto dm, the first integral is simply Icom, the rotational inertia of the
body about an axis through its center of mass. Inspection of Fig. 10-12 shows that
the last term in Eq. 10-37 is Mh2, where Mis the body’s total mass. Thus,
Eq. 10-37 reduces to Eq. 10-36, which is the relation that we set out to prove.
Checkpoint 5
The figure shows a book-like object (one side is
longer than the other) and four choices of rotation
axes,all perpendicular to the face of the object.
Rank the choices according to the rotational inertia
of the object about the axis,greatest first.
(
1
)(
2
)(
3
)(
4
)
left and Lfor the particle on the right. Now Eq. 10-33
gives us
Im(0)2mL2mL2. (Answer)
Second technique: Because we already know Icom about an
axis through the center of mass and because the axis here is
parallel to that “com axis, we can apply the parallel-axis
theorem (Eq. 10-36).We find
(Answer)mL2.
IIcom Mh21
2mL2(2m)(1
2L)2
Sample Problem 10.06 Rotational inertia of a two-particle system
Figure 10-13ashows a rigid body consisting of two particles of
mass mconnected by a rod of length Land negligible mass.
(a) What is the rotational inertia Icom about an axis through the
center of mass,perpendicular to the rod as shown?
KEY IDEA
Because we have only two particles with mass, we can find
the body’s rotational inertia Icom by using Eq. 10-33 rather
than by integration. That is, we find the rotational inertia of
each particle and then just add the results.
Calculations: For the two particles, each at perpendicular
distance from the rotation axis, we have
(Answer)
(b) What is the rotational inertia Iof the body about an axis
through the left end of the rod and parallel to the first axis
(Fig. 10-13b)?
KEY IDEAS
This situation is simple enough that we can find Iusing
either of two techniques. The first is similar to the one used
in part (a). The other, more powerful one is to apply the
parallel-axis theorem.
First technique: We calculate Ias in part (a), except here
the perpendicular distance riis zero for the particle on the
1
2mL2.
I
miri
2(m)(1
2L)2(m)(1
2L)2
1
2L
Additional examples, video, and practice available at WileyPLUS
m m
(a)
LL
com
Rotat
i
on ax
i
s
through
center of mass
mm
(b)
L
com
Rotation axis through
end of rod
1
__
2
1
__
2
Here the rotation axis is through the com.
Here it has been shifted from the com
without changing the orientation. We
can use the parallel-axis theorem.
Figure 10-13 A rigid body consisting of two particles of mass m
joined by a rod of negligible mass.
276 CHAPTER 10 ROTATION
Sample Problem 10.07 Rotational inertia of a uniform rod, integration
Figure 10-14 shows a thin, uniform rod of mass Mand length
L, on an xaxis with the origin at the rod’s center.
(a) What is the rotational inertia of the rod about the
perpendicular rotation axis through the center?
KEY IDEAS
(1) The rod consists of a huge number of particles at a great
many different distances from the rotation axis. We certainly
don’t want to sum their rotational inertias individually. So, we
first write a general expression for the rotational inertia of a
mass element dm at distance rfrom the rotation axis: r2dm.
(2) Then we sum all such rotational inertias by integrating the
expression (rather than adding them up one by one). From
Eq. 10-35,we write
(10-38)
(3) Because the rod is uniform and the rotation axis is at the
center, we are actually calculating the rotational inertia Icom
about the center of mass.
Calculations: We want to integrate with respect to coordinate
x(not mass mas indicated in the integral), so we must relate
the mass dm of an element of the rod to its length dx along the
rod. (Such an element is shown in Fig. 10-14.) Because the rod
is uniform, the ratio of mass to length is the same for all the el-
ements and for the rod as a whole.Thus, we can write
or dm M
Ldx.
element’s mass dm
element’s length dx rod’s mass M
rod’s length L
Ir2dm.
Figure 10-14 A uniform rod of length L
and mass M. An element of mass dm
and length dx is represented.
A
We can now substitute this result for dm and xfor rin
Eq. 10-38.Then we integrate from end to end of the rod (from
xL/2 to xL/2) to include all the elements.We find
(Answer)
(b) What is the rod’s rotational inertia Iabout a new rotation
axis that is perpendicular to the rod and through the left end?
KEY IDEAS
We can find Iby shifting the origin of the xaxis to the left end
of the rod and then integrating from to .However,
here we shall use a more powerful (and easier) technique by
applying the parallel-axis theorem (Eq. 10-36), in which we
shift the rotation axis without changing its orientation.
Calculations: If we place the axis at the rod’s end so that it
is parallel to the axis through the center of mass, then we
can use the parallel-axis theorem (Eq. 10-36). We know
from part (a) that Icom is . From Fig. 10-14, the perpen-
dicular distance hbetween the new rotation axis and the
center of mass is . Equation 10-36 then gives us
(Answer)
Actually, this result holds for any axis through the left
or right end that is perpendicular to the rod.
1
3ML2.
IIcom Mh21
12 ML2(M)(1
2L)2
1
2L
1
12 ML2
xLx 0
1
12 ML2.
M
3L
x3
L/2
L/2
M
3L
L
2
3
L
2
3
IxL/2
xL/2 x2
M
L
dx
Additional examples, video, and practice available at WileyPLUS
x
Rotation
axis
L
__
2
L
__
2
com M
We want the
rotational inertia.
x
Rotation
axis
xdm
dx
First, pick any tiny element
and write its rotational
inertia as x2dm.
x
x=
Rotation
axis
Leftmost Rightmost
L
__
2x=L
__
2
Then, using integration, add up
the rotational inertias for all of
the elements, from leftmost to
rightmost.
277
10-6 TORQUE
KEY IDEA
The released energy was equal to the rotational kinetic en-
ergy Kof the rotor just as it reached the angular speed of
14 000 rev/min.
Calculations: We can find Kwith Eq. 10-34 , but
first we need an expression for the rotational inertia I. Because
the rotor was a disk that rotated like a merry-go-round, Iis
given in Table 10-2c.Thus,
The angular speed of the rotor was
Then, with Eq. 10-34, we find the (huge) energy release:
(Answer)2.1 107 J.
K1
2Iv21
2(19.64 kgm2)(1.466 103 rad/s)2
1.466 103 rad/s.
v(14 000 rev/min)(2p rad/rev)
1 min
60 s
I1
2MR21
2(272 kg)(0.38 m)219.64 kgm2.
(I1
2MR2)
(K1
2Iv2)
Sample Problem 10.08 Rotational kinetic energy, spin test explosion
Large machine components that undergo prolonged, high-
speed rotation are first examined for the possibility of fail-
ure in a spin test system. In this system, a component is spun
up (brought up to high speed) while inside a cylindrical
arrangement of lead bricks and containment liner, all within
a steel shell that is closed by a lid clamped into place. If the
rotation causes the component to shatter, the soft lead
bricks are supposed to catch the pieces for later analysis.
In 1985, Test Devices, Inc. (www.testdevices.com) was spin
testing a sample of a solid steel rotor (a disk) of mass M
272 kg and radius R38.0 cm. When the sample reached
an angular speed vof 14 000 rev/min, the test engineers
heard a dull thump from the test system, which was
located one floor down and one room over from them.
Investigating, they found that lead bricks had been thrown
out in the hallway leading to the test room, a door to the
room had been hurled into the adjacent parking lot, one
lead brick had shot from the test site through the wall of a
neighbor’s kitchen, the structural beams of the test build-
ing had been damaged, the concrete floor beneath the
spin chamber had been shoved downward by about 0.5
cm, and the 900 kg lid had been blown upward through
the ceiling and had then crashed back onto the test equip-
ment (Fig. 10-15). The exploding pieces had not pene-
trated the room of the test engineers only by luck.
How much energy was released in the explosion of the
rotor?
Figure 10-15 Some of the
destruction caused by
the explosion of a rap-
idly rotating steel disk.
Courtesy Test Devices, Inc.
10-6 TORQUE
After reading this module, you should be able to . . .
10.23 Identify that a torque on a body involves a force and a
position vector, which extends from a rotation axis to the
point where the force is applied.
10.24 Calculate the torque by using (a) the angle between
the position vector and the force vector, (b) the line of ac-
tion and the moment arm of the force, and (c) the force
component perpendicular to the position vector.
10.25 Identify that a rotation axis must always be specified to
calculate a torque.
10.26 Identify that a torque is assigned a positive or negative
sign depending on the direction it tends to make the body
rotate about a specified rotation axis: “clocks are negative.
10.27 When more than one torque acts on a body about a
rotation axis, calculate the net torque.
Learning Objectives
Torque is a turning or twisting action on a body about a
rotation axis due to a force . If is exerted at a point given
by the position vector relative to the axis, then the magni-
tude of the torque is
where Ftis the component of perpendicular to and
fis the angle between and . The quantity is the r
F
:
r
:
r
:
F
:
trF
trFrF sin f,
r
:
F
:
F
:perpendicular distance between the rotation axis and
an extended line running through the vector. This line
is called the line of action of , and is called the
moment arm of . Similarly, ris the moment arm of Ft.
The SI unit of torque is the newton-meter (Nm). A
torque tis positive if it tends to rotate a body at rest
counterclockwise and negative if it tends to rotate the
body clockwise.
F
:r
F
:F
:
Key Ideas
Additional examples, video, and practice available at WileyPLUS
278 CHAPTER 10 ROTATION
Checkpoint 6
The figure shows an overhead view of a meter stick that can pivot about the dot at the position
marked 20 (for 20 cm).All five forces on the stick are horizontal and have the same magnitude.
Rank the forces according to the magnitude of the torque they produce, greatest first.
0 20 40
P
i
vot po
i
nt
100
F1
F2F3
F4
F5
Figure 10-16 (a) A force acts on a rigid
body, with a rotation axis perpendicular to
the page. The torque can be found with
(a) angle f,(b) tangential force compo-
nent Ft, or (c) moment arm .r
F
:
(a)
(b)
(c)
O
P
φ
FrFt
Rotation
axis
F
r
O
P
φ
Rotation
axis
φ
Line of
action of F
r
Moment arm
of F
F
r
O
P
φ
Rotation
axis
F
r
The torque due to this force
causes rotation around this axis
(which extends out toward you).
You calculate the same torque by
using this moment arm distance
and the full force magnitude.
But actually only the tangential
component of the force causes
the rotation.
magnitude FtFsin f.This component does cause rotation.
Calculating Torques. The ability of to rotate the body depends not only
on the magnitude of its tangential component Ft, but also on just how far from O
the force is applied. To include both these factors, we define a quantity called
torque tas the product of the two factors and write it as
t(r)(Fsin f). (10-39)
Two equivalent ways of computing the torque are
t(r)(Fsin f)rFt(10-40)
and (10-41)
where is the perpendicular distance between the rotation axis at Oand an extendedr
t(r sin f)(F)rF,
F
:
Torque
A doorknob is located as far as possible from the door’s hinge line for a good rea-
son. If you want to open a heavy door, you must certainly apply a force, but that
is not enough.Where you apply that force and in what direction you push are also
important. If you apply your force nearer to the hinge line than the knob, or at
any angle other than 90to the plane of the door, you must use a greater force
than if you apply the force at the knob and perpendicular to the door’s plane.
Figure 10-16ashows a cross section of a body that is free to rotate about an
axis passing through Oand perpendicular to the cross section. A force is
applied at point P, whose position relative to Ois defined by a position vector .
The directions of vectors and make an angle fwith each other. (For simplic-
ity, we consider only forces that have no component parallel to the rotation axis;
thus, is in the plane of the page.)
To determine how results in a rotation of the body around the rotation
axis, we resolve into two components (Fig. 10-16b). One component, called the
radial component Fr, points along . This component does not cause rotation,
because it acts along a line that extends through O. (If you pull on a door par-
allel to the plane of the door, you do not rotate the door.) The other compo-
nent of , called the tangential component Ft, is perpendicular to and hasr
:
F
:
r
:
F
:F
:
F
:
r
:
F
:r
:
F
:
line running through the vector (Fig. 10-16c). This extended line is called the line
of action of , and is called the moment arm of . Figure 10-16bshows that we
can describe r,the magnitude of ,as being the moment arm of the force component Ft.
Torque, which comes from the Latin word meaning “to twist, may be loosely
identified as the turning or twisting action of the force .When you apply a force
to an objectsuch as a screwdriver or torque wrenchwith the purpose of turn-
ing that object, you are applying a torque. The SI unit of torque is the newton-
meter (Nm). Caution: The newton-meter is also the unit of work. Torque and
work, however, are quite different quantities and must not be confused. Work is
often expressed in joules (1 J 1Nm), but torque never is.
Clocks Are Negative. In Chapter 11 we shall use vector notation for torques,
but here, with rotation around a single axis, we use only an algebraic sign. If a
torque would cause counterclockwise rotation, it is positive. If it would cause
clockwise rotation, it is negative. (The phrase “clocks are negative” from Module
10-1 still works.)
Torques obey the superposition principle that we discussed in Chapter 5 for
forces:When several torques act on a body, the net torque (or resultant torque) is
the sum of the individual torques.The symbol for net torque is tnet.
F
:
r
:
F
:
r
F
:F
:
279
10-7 NEWTON’S SECOND LAW FOR ROTATION
10-7 NEWTON’S SECOND LAW FOR ROTATION
After reading this module, you should be able to . . .
10.28 Apply Newton’s second law for rotation to relate the
net torque on a body to the body’s rotational inertia and
rotational acceleration, all calculated relative to a specified
rotation axis.
The rotational analog of Newton’s second law is
tnet Ia,
where tnet is the net torque acting on a particle or rigid body,
Iis the rotational inertia of the particle or body about the
rotation axis, and ais the resulting angular acceleration about
that axis.
Learning Objective
Key Idea
Newton’s Second Law for Rotation
A torque can cause rotation of a rigid body, as when you use a torque to rotate
a door. Here we want to relate the net torque tnet on a rigid body to the angular
acceleration athat torque causes about a rotation axis. We do so by analogy with
Newton’s second law (Fnet ma) for the acceleration aof a body of mass mdue
to a net force Fnet along a coordinate axis.We replace Fnet with tnet,mwith I, and a
with ain radian measure, writing
tnet Ia(Newton’s second law for rotation). (10-42)
Proof of Equation 10-42
We prove Eq. 10-42 by first considering the simple situation shown in Fig. 10-17.
The rigid body there consists of a particle of mass mon one end of a massless rod
of length r.The rod can move only by rotating about its other end, around a rota-
tion axis (an axle) that is perpendicular to the plane of the page.Thus, the particle
can move only in a circular path that has the rotation axis at its center.
A force acts on the particle. However, because the particle can move
only along the circular path, only the tangential component Ftof the force (the
component that is tangent to the circular path) can accelerate the particle along
the path. We can relate Ftto the particle’s tangential acceleration atalong the
path with Newton’s second law, writing
Ftmat.
The torque acting on the particle is, from Eq. 10-40,
tFtrmatr.
From Eq. 10-22 (atar) we can write this as
tm(ar)r(mr2)a. (10-43)
The quantity in parentheses on the right is the rotational inertia of the particle
about the rotation axis (see Eq. 10-33, but here we have only a single particle).
Thus, using Ifor the rotational inertia, Eq. 10-43 reduces to
tIa(radian measure). (10-44)
If more than one force is applied to the particle, Eq. 10-44 becomes
tnet Ia(radian measure), (10-45)
which we set out to prove.We can extend this equation to any rigid body rotating
about a fixed axis, because any such body can always be analyzed as an assembly
of single particles.
F
:
Figure 10-17 A simple rigid body, free to
rotate about an axis through O, consists of
a particle of mass mfastened to the end of
a rod of length rand negligible mass. An
applied force causes the body to rotate.F
:
Ox
y
Rod
θ
Rotation axis
r
m
Fr
Ft
φ
F
The torque due to the tangential
component of the force causes
an angular acceleration around
the rotation axis.
280 CHAPTER 10 ROTATION
Additional examples, video, and practice available at WileyPLUS
KEY IDEA
Because the moment arm for is no longer zero, the torqueF
:
g
Checkpoint 7
The figure shows an overhead view of a meter stick that can pivot about the point indicated, which is
to the left of the stick’s midpoint.Two horizontal forces, and ,are applied to the stick.Only is
shown. Force is perpendicular to the stick and is applied at the right end. If the stick is not to turn,
(a) what should be the direction of , and (b) should F2be greater than,less than, or equal to F1?F
:
2
F
:
2
F
:
1
F
:
2
F
:
1
F
1
Pivot point
Sample Problem 10.09 Using Newton’s second law for rotation in a basic judo hip throw
To throw an 80 kg opponent with a basic judo hip throw, you
intend to pull his uniform with a force and a moment arm
d10.30 m from a pivot point (rotation axis) on your right
hip (Fig. 10-18). You wish to rotate him about the pivot
point with an angular acceleration aof 6.0 rad/s2—that is,
with an angular acceleration that is clockwise in the figure.
Assume that his rotational inertia Irelative to the pivot
point is 15 kgm2.
(a) What must the magnitude of be if, before you throw
him, you bend your opponent forward to bring his center of
mass to your hip (Fig.10-18a)?
KEY IDEA
We can relate your pull on your opponent to the given an-
gular acceleration avia Newton’s second law for rotation
(tnet Ia).
Calculations: As his feet leave the floor, we can assume that
only three forces act on him: your pull , a force on him
from you at the pivot point (this force is not indicated in Fig.
10-18), and the gravitational force .To use tnet Ia, we need
the corresponding three torques, each about the pivot point.
From Eq. 10-41 (tF), the torque due to your pull F
:
r
F
:
g
N
:
F
:
F
:
F
:
F
:
Figure 10-18 A judo hip throw (a) correctly executed and (b) incor-
rectly executed.
Opponent's
center of
mass
Moment arm d1
of your pull
Pivot
on hip
Moment arm d2
of gravitational
force on
opponent
Moment
arm d1
of your pull
Fg
Fg
(a) (b)
F
F
is equal to F, where is the moment arm and the
sign indicates the clockwise rotation this torque tends to
cause. The torque due to is zero, because acts at theN
:
N
:
r
d1
d1
pivot point and thus has moment arm 0.
To evaluate the torque due to , we can assume that
acts at your opponent’s center of mass. With the center of
mass at the pivot point, has moment arm 0 and thusr
F
:
g
F
:
g
F
:
g
r
ponent is due to your pull , and we can write tnet Iaas
d1FIa.
We then find
300 N. (Answer)
(b) What must the magnitude of be if your opponent
remains upright before you throw him, so that has a mo-
ment arm d20.12 m (Fig.10-18b)?
F
:
g
F
:
FIa
d1
(15 kgm2)(6.0 rad/s2)
0.30 m
F
:
the torque due to is zero. So, the only torque on your op-
F
:
g
due to is now equal to d2mg and is positive because the
torque attempts counterclockwise rotation.
Calculations: Now we write tnet Iaas
d1Fd2mg Ia,
which gives
From (a), we know that the first term on the right is equal to
300 N. Substituting this and the given data, we have
613.6 N 610 N. (Answer)
The results indicate that you will have to pull much harder if
you do not initially bend your opponent to bring his center
of mass to your hip. A good judo fighter knows this lesson
from physics. Indeed, physics is the basis of most of the mar-
tial arts, figured out after countless hours of trial and error
over the centuries.
F300 N (0.12 m)(80 kg)(9.8 m/s2)
0.30 m
FIa
d1
d2mg
d1
.
F
:
g
281
10-7 NEWTON’S SECOND LAW FOR ROTATION
Sample Problem 10.10 Newton’s second law, rotation, torque, disk
Figure 10-19ashows a uniform disk, with mass M2.5 kg
and radius R20 cm, mounted on a fixed horizontal axle.
A block with mass m1.2 kg hangs from a massless cord that
is wrapped around the rim of the disk. Find the acceleration of
the falling block, the angular acceleration of the disk, and the
tension in the cord.The cord does not slip, and there is no fric-
tion at the axle.
KEY IDEAS
(1) Taking the block as a system, we can relate its accelera-
tion ato the forces acting on it with Newton’s second law
( ). (2) Taking the disk as a system, we can relate
its angular acceleration ato the torque acting on it with
Newton’s second law for rotation (tnet Ia). (3) To combine
the motions of block and disk, we use the fact that the linear
acceleration aof the block and the (tangential) linear accel-
eration of the disk rim are equal. (To avoid confusion
about signs, let’s work with acceleration magnitudes and
explicit algebraic signs.)
Forces on block: The forces are shown in the block’s free-
body diagram in Fig. 10-19b: The force from the cord is ,
and the gravitational force is , of magnitude mg. We can
now write Newton’s second law for components along a ver-
tical yaxis (Fnet,ymay) as
Tmg m(a), (10-46)
where ais the magnitude of the acceleration (down the y
axis). However, we cannot solve this equation for abecause
it also contains the unknown T.
Torque on disk: Previously, when we got stuck on the yaxis,
we switched to the xaxis. Here, we switch to the rotation of
the disk and use Newton’s second law in angular form. To
calculate the torques and the rotational inertia I, we take
the rotation axis to be perpendicular to the disk and through
its center,at point Oin Fig. 10-19c.
The torques are then given by Eq. 10-40 (trFt). The
gravitational force on the disk and the force on the disk from
the axle both act at the center of the disk and thus at distance
r0, so their torques are zero.The force on the disk due to
the cord acts at distance rRand is tangent to the rim of the
disk. Therefore, its torque is RT, negative because the
torque rotates the disk clockwise from rest. Let abe the mag-
nitude of the negative (clockwise) angular acceleration. From
Table 10-2c, the rotational inertia Iof the disk is . Thus
we can write the general equation tnet Iaas
(10-47)RT 1
2MR2(a).
1
2MR2
T
:
F
:
g
T
:
at
F
:
net m:
a
This equation seems useless because it has two
unknowns, aand T, neither of which is the desired a.
However, mustering physics courage, we can make it useful
with this fact: Because the cord does not slip, the magnitude
aof the block’s linear acceleration and the magnitude atof
the (tangential) linear acceleration of the rim of the disk are
equal. Then, by Eq. 10-22 (atar) we see that here a
a/R. Substituting this in Eq. 10-47 yields
(10-48)
Combining results: Combining Eqs. 10-46 and 10-48 leads to
. (Answer)
We then use Eq. 10-48 to find T:
(Answer)
As we should expect, acceleration aof the falling block is less
than g, and tension Tin the cord (6.0 N) is less than the
gravitational force on the hanging block (mg 11.8 N).
We see also that aand Tdepend on the mass of the disk but
not on its radius.
As a check, we note that the formulas derived above
predict agand T0 for the case of a massless disk (M
0). This is what we would expect; the block simply falls as a
free body. From Eq. 10-22, the magnitude of the angular ac-
celeration of the disk is
(Answer)aa
R4.8 m/s2
0.20 m 24 rad/s2.
6.0 N.
T1
2Ma 1
2(2.5 kg)(4.8 m/s2)
4.8 m/s2
ag2m
M2m(9.8 m/s2)(2)(1.2 kg)
2.5 kg (2)(1.2 kg)
T1
2Ma.
m
M
MR
O
Fg
(b)(a)
(c)
m
T
T
The torque due to the
cord's pull on the rim
causes an angular
acceleration of the disk.
These two forces
determine the block's
(linear) acceleration.
We need to relate
those two
accelerations.
y
Figure 10-19 (a) The falling block causes the disk to rotate. (b) A
free-body diagram for the block. (c) An incomplete free-body
diagram for the disk.
Additional examples, video, and practice available at WileyPLUS
282 CHAPTER 10 ROTATION
Work and Rotational Kinetic Energy
As we discussed in Chapter 7, when a force Fcauses a rigid body of mass mto ac-
celerate along a coordinate axis, the force does work Won the body. Thus, the
body’s kinetic energy can change. Suppose it is the only energy of the(K1
2mv2)
10-8 WORK AND ROTATIONAL KINETIC ENERGY
After reading this module, you should be able to . . .
10.29 Calculate the work done by a torque acting on a rotat-
ing body by integrating the torque with respect to the an-
gle of rotation.
10.30 Apply the work–kinetic energy theorem to relate the
work done by a torque to the resulting change in the rota-
tional kinetic energy of the body.
10.31 Calculate the work done by a constant torque by relat-
ing the work to the angle through which the body rotates.
10.32 Calculate the power of a torque by finding the rate at
which work is done.
10.33 Calculate the power of a torque at any given instant by
relating it to the torque and the angular velocity at that instant.
The equations used for calculating work and power in rota-
tional motion correspond to equations used for translational
motion and are
and PdW
dt tv.
Wu
f
u
i
t
du
When tis constant, the integral reduces to
Wt(ufui).
The form of the work kinetic energy theorem used for
rotating bodies is
KKfKi1
2Ivf
21
2
vi
2W.
Learning Objectives
Key Ideas
body that changes.Then we relate the change Kin kinetic energy to the work W
with the workkinetic energy theorem (Eq. 7-10), writing
(workkinetic energy theorem). (10-49)
For motion confined to an xaxis, we can calculate the work with Eq. 7-32,
(work, one-dimensional motion). (10-50)
This reduces to WFd when Fis constant and the body’s displacement is d.
The rate at which the work is done is the power, which we can find with Eqs. 7-43
and 7-48,
(power,one-dimensional motion). (10-51)
Now let us consider a rotational situation that is similar. When a torque
accelerates a rigid body in rotation about a fixed axis, the torque does work W
on the body. Therefore, the body’s rotational kinetic energy can
change. Suppose that it is the only energy of the body that changes. Then we
can still relate the change Kin kinetic energy to the work Wwith the
workkinetic energy theorem, except now the kinetic energy is a rotational
kinetic energy:
(workkinetic energy theorem). (10-52)
Here, Iis the rotational inertia of the body about the fixed axis and viand vfare
the angular speeds of the body before and after the work is done.
KKfKi1
2Ivf
21
2
vi
2W
(K1
2I
2)
PdW
dt Fv
Wxf
xi
Fdx
KKfKi1
2mvf
21
2mvi
2W
283
10-8 WORK AND ROTATIONAL KINETIC ENERGY
Also, we can calculate the work with a rotational equivalent of Eq. 10-50,
(work, rotation about fixed axis), (10-53)
where tis the torque doing the work W, and uiand ufare the body’s angular
positions before and after the work is done, respectively. When tis constant,
Eq. 10-53 reduces to
Wt(ufui)(work, constant torque). (10-54)
The rate at which the work is done is the power, which we can find with the rota-
tional equivalent of Eq. 10-51,
(power,rotation about fixed axis). (10-55)
Table 10-3 summarizes the equations that apply to the rotation of a rigid body
about a fixed axis and the corresponding equations for translational motion.
Proof of Eqs. 10-52 through 10-55
Let us again consider the situation of Fig. 10-17, in which force rotates a rigid
body consisting of a single particle of mass mfastened to the end of a massless
rod. During the rotation, force does work on the body. Let us assume that the
only energy of the body that is changed by is the kinetic energy. Then we can
apply the workkinetic energy theorem of Eq. 10-49:
KKfKiW. (10-56)
Using and Eq. 10-18 (vvr), we can rewrite Eq. 10-56 as
(10-57)
From Eq. 10-33, the rotational inertia for this one-particle body is Imr2.
Substituting this into Eq. 10-57 yields
which is Eq. 10-52.We derived it for a rigid body with one particle, but it holds for
any rigid body rotated about a fixed axis.
We next relate the work Wdone on the body in Fig. 10-17 to the torque t
on the body due to force . When the particle moves a distance ds along itsF
:
K1
2Ivf
21
2
vi
2W,
K1
2mr2vf
21
2mr2vi
2W.
K1
2mv2
F
:
F
:
F
:
PdW
dt tv
Wuf
ui
tdu
Table 10-3 Some Corresponding Relations for Translational and Rotational Motion
Pure Translation (Fixed Direction) Pure Rotation (Fixed Axis)
Position xAngular position u
Velocity vdx/dt Angular velocity vdu/dt
Acceleration adv/dt Angular acceleration adv/dt
Mass mRotational inertia I
Newton’s second law Fnet ma Newton’s second law tnet Ia
Work WF dx Work Wtdu
Kinetic energy Kinetic energy K1
2Iv2
K1
2mv2
Power (constant force) PFv Power (constant torque) Ptv
Workkinetic energy theorem WKWorkkinetic energy theorem WK
284 CHAPTER 10 ROTATION
Sample Problem 10.11 Work, rotational kinetic energy, torque, disk
Let the disk in Fig. 10-19 start from rest at time t0 and
also let the tension in the massless cord be 6.0 N and the an-
gular acceleration of the disk be 24 rad/s2.What is its rota-
tional kinetic energy Kat t2.5 s?
KEY IDEA
We can find Kwith Eq. 10-34 We already know(K1
2Iv2).
Calculations: First, we relate the change in the kinetic
energy of the disk to the net work Wdone on the disk, using
the workkinetic energy theorem of Eq. 10-52 (KfKiW).
With Ksubstituted for Kfand 0 for Ki,we get
KKiW0WW. (10-60)
Next we want to find the work W. We can relate Wto
the torques acting on the disk with Eq. 10-53 or 10-54. The
only torque causing angular acceleration and doing work is
the torque due to force on the disk from the cord, which isT
:
that , but we do not yet know vat t2.5 s.
However, because the angular acceleration ahas the con-
stant value of 24 rad/s2, we can apply the equations for
constant angular acceleration in Table 10-1.
Calculations: Because we want vand know aand v0(0),
we use Eq. 10-12:
vv0at0atat.
Substituting vatand into Eq.10-34, we find
(Answer)
KEY IDEA
We can also get this answer by finding the disk’s kinetic
energy from the work done on the disk.
90 J.
1
4(2.5 kg)[(0.20 m)(24 rad/s2)(2.5 s)]2
K1
2Iv21
2(1
2MR2)(at)21
4M(Rat)2
I1
2MR2
I1
2MR2
Additional examples, video, and practice available at WileyPLUS
circular path, only the tangential component Ftof the force accelerates the parti-
cle along the path. Therefore, only Ftdoes work on the particle. We write that
work dW as Ftds. However, we can replace ds with r du, where duis the angle
through which the particle moves.Thus we have
dW Ftr du. (10-58)
From Eq. 10-40, we see that the product Ftris equal to the torque t, so we can
rewrite Eq. 10-58 as
dW tdu. (10-59)
The work done during a finite angular displacement from uito ufis then
which is Eq. 10-53. It holds for any rigid body rotating about a fixed axis.
Equation 10-54 comes directly from Eq. 10-53.
We can find the power Pfor rotational motion from Eq. 10-59:
which is Eq. 10-55.
PdW
dt t
du
dt tv,
Wuf
ui
tdu,
equal to TR. Because ais constant, this torque also must
be constant.Thus, we can use Eq. 10-54 to write
Wt(ufui)TR(ufui). (10-61)
Because ais constant, we can use Eq. 10-13 to find
ufui.With vi0, we have
.
Now we substitute this into Eq. 10-61 and then substitute the
result into Eq. 10-60. Inserting the given values T6.0 N
and a24 rad/s2, we have
(Answer)90 J.

1
2(6.0 N)(0.20 m)(24 rad/s2)(2.5 s)2
KWTR(ufui)TR(1
2at2)
1
2TRat2
ufuivit1
2at201
2at21
2at2
285
REVIEW & SUMMARY
Angular Position To describe the rotation of a rigid body about
a fixed axis, called the rotation axis, we assume a reference line is
fixed in the body, perpendicular to that axis and rotating with the
body.We measure the angular position uof this line relative to a fixed
direction.When uis measured in radians,
(radian measure), (10-1)
where sis the arc length of a circular path of radius rand angle u.
Radian measure is related to angle measure in revolutions and de-
grees by 1 rev 3602prad. (10-2)
Angular Displacement A body that rotates about a rotation
axis,changing its angular position from u1to u2, undergoes an angu-
lar displacement uu2u1, (10-4)
where uis positive for counterclockwise rotation and negative for
clockwise rotation.
Angular Velocity and Speed If a body rotates through an
angular displacement uin a time interval t, its average angular
velocity vavg is
(10-5)
The (instantaneous) angular velocity vof the body is
(10-6)
Both vavg and vare vectors, with directions given by the right-hand
rule of Fig. 10-6. They are positive for counterclockwise rotation
and negative for clockwise rotation. The magnitude of the body’s
angular velocity is the angular speed.
Angular Acceleration If the angular velocity of a body
changes from v1to v2in a time interval tt2t1, the average
angular acceleration aavg of the body is
(10-7)
The (instantaneous) angular acceleration aof the body is
(10-8)
Both aavg and aare vectors.
The Kinematic Equations for Constant Angular Accel-
eration Constant angular acceleration (aconstant) is an im-
portant special case of rotational motion. The appropriate kine-
matic equations,given in Table 10-1, are
vv0at, (10-12)
(10-13)
(10-14)
(10-15)
(10-16)
Linear and Angular Variables Related A point in a rigid
rotating body, at a perpendicular distance r from the rotation axis,
uu0vt1
2at2.
uu01
2(v0v)t,
v2v0
22a(uu0),
uu0v0t1
2at2,
adv
dt .
aavg v2v1
t2t1
v
t.
vdu
dt .
vavg u
t.
us
r
Review & Summary
moves in a circle with radius r. If the body rotates through an angle u,
the point moves along an arc with length sgiven by
sur(radian measure), (10-17)
where uis in radians.
The linear velocity of the point is tangent to the circle; the
point’s linear speed vis given by
vvr(radian measure), (10-18)
where vis the angular speed (in radians per second) of the body.
The linear acceleration of the point has both tangential and
radial components.The tangential component is
atar(radian measure), (10-22)
where ais the magnitude of the angular acceleration (in radians
per second-squared) of the body.The radial component of is
(radian measure). (10-23)
If the point moves in uniform circular motion, the period Tof
the motion for the point and the body is
(radian measure). (10-19, 10-20)
Rotational Kinetic Energy and Rotational Inertia The ki-
netic energy Kof a rigid body rotating about a fixed axis is given by
(radian measure), (10-34)
in which Iis the rotational inertia of the body, defined as
(10-33)
for a system of discrete particles and defined as
(10-35)
for a body with continuously distributed mass. The rand riin these
expressions represent the perpendicular distance from the axis of
rotation to each mass element in the body, and the integration is car-
ried out over the entire body so as to include every mass element.
The Parallel-Axis Theorem The parallel-axis theorem relates
the rotational inertia Iof a body about any axis to that of the same
body about a parallel axis through the center of mass:
IIcom Mh2. (10-36)
Here his the perpendicular distance between the two axes, and
Icom is the rotational inertia of the body about the axis through the
com. We can describe has being the distance the actual rotation
axis has been shifted from the rotation axis through the com.
Torque Torque is a turning or twisting action on a body about a ro-
tation axis due to a force . If is exerted at a point given by the po-
sition vector relative to the axis, then the magnitude of the torque is
(10-40, 10-41, 10-39)
where Ftis the component of perpendicular to and fis the an-
gle between and .The quantity is the perpendicular distance
between the rotation axis and an extended line running through
the vector. This line is called the line of action of , and is
called the moment arm of . Similarly, ris the moment arm of Ft.F
:r
F
:
F
:
r
F
:
r
:
r
:
F
:
trF
trFrF sin f,
r
:
F
:
F
:
Ir2dm
I
miri
2
K1
2Iv2
T2pr
v2p
v
arv2
rv2r
a
:
a
:
v
:
8Figure 10-25bshows an overhead view of a horizontal bar that
is rotated about the pivot point by two horizontal forces, and ,
with at angle fto the bar. Rank the following values of faccord-
ing to the magnitude of the angular acceleration of the bar, greatest
first: 90,70, and 110.
9Figure 10-26 shows a uniform metal plate
that had been square before 25% of it was
snipped off. Three lettered points are indicated.
Rank them according to the rotational inertia of
the plate around a perpendicular axis through
them, greatest first.
F
:
2
F
:
2
F
:
1
286 CHAPTER 10 ROTATION
angles during the rotation, which is
counterclockwise and at a constant
rate. However, we are to decrease the
angle uof without changing the
magnitude of . (a) To keep the an-
gular speed constant, should we in-
crease, decrease, or maintain the mag-
nitude of ? Do forces (b) and (c)
tend to rotate the disk clockwise or
counterclockwise?
6In the overhead view of Fig. 10-24,
five forces of the same magnitude act
on a strange merry-go-round; it is a
square that can rotate about point P,at
midlength along one of the edges.
Rank the forces according to the mag-
nitude of the torque they create about
point P,greatest first.
7Figure 10-25ais an overhead view
of a horizontal bar that can pivot; two horizontal forces act on the
bar, but it is stationary. If the angle between the bar and is nowF
:
2
F
:
2
F
:
1
F
:
2
F
:
1
F
:
1
1Figure 10-20 is a graph of the an-
gular velocity versus time for a disk
rotating like a merry-go-round. For a
point on the disk rim, rank the in-
stants a,b,c, and daccording to the
magnitude of the (a) tangential and
(b) radial acceleration, greatest first.
2Figure 10-21 shows plots of angu-
lar position uversus time tfor three
cases in which a disk is rotated like a
merry-go-round. In each case, the ro-
tation direction changes at a certain
angular position uchange. (a) For each
case, determine whether uchange is
clockwise or counterclockwise from
u0, or whether it is at u0. For
each case, determine (b) whether
vis zero before, after, or at t0
and (c) whether ais positive,negative,or zero.
3A force is applied to the rim of a disk that can rotate like
a merry-go-round, so as to change its angular velocity. Its initial
and final angular velocities, respectively, for four situations are:
(a) 2 rad/s, 5 rad/s; (b) 2 rad/s, 5 rad/s; (c) 2 rad/s, 5 rad/s; and
(d) 2 rad/s, 5 rad/s. Rank the situations according to the work
done by the torque due to the force, greatest first.
4Figure 10-22bis a graph of the angular position of the rotating
disk of Fig. 10-22a. Is the angular velocity of the disk positive,nega-
tive, or zero at (a) t1 s, (b) t2 s, and (c) t3 s? (d) Is the an-
gular acceleration positive or negative?
Questions
ω
t
abcd
Figure 10-20 Question 1.
1
2
3
0
–90°
90°
θ
t
Figure 10-21 Question 2.
Rotation axis
t(s)
(rad)
θ
1 2 3
(a) (b)
Figure 10-22 Question 4.
F1
θ
F2
Figure 10-23 Question 5.
Figure 10-24 Question 6.
F5
F4
F3
F2
F1P
Pivot point
F1F2Pivot point
(
a
)(
b
)
φ
F1
F2
a
b
c
Figure 10-26
Question 9.
Figure 10-25 Questions 7 and 8.
The SI unit of torque is the newton-meter (Nm). A torque t
is positive if it tends to rotate a body at rest counterclockwise and
negative if it tends to rotate the body clockwise.
Newton’s Second Law in Angular Form The rotational
analog of Newton’s second law is
tnet Ia, (10-45)
where tnet is the net torque acting on a particle or rigid body, Iis the ro-
tational inertia of the particle or body about the rotation axis, and ais
the resulting angular acceleration about that axis.
Work and Rotational Kinetic Energy The equations used
for calculating work and power in rotational motion correspond to
equations used for translational motion and are
(10-53)
and (10-55)
When tis constant,Eq. 10-53 reduces to
Wt(ufui). (10-54)
The form of the work kinetic energy theorem used for rotating
bodies is
(10-52)KKfKi1
2Ivf
21
2
vi
2W.
PdW
dt tv.
Wu
f
u
i
t
du
5In Fig.10-23, two forces and act on a disk that turns about
its center like a merry-go-round.The forces maintain the indicated
F
:
2
F
:
1
decreased from 90and the bar is still not to turn, should F2be
made larger,made smaller,or left the same?
287
PROBLEMS
10 Figure 10-27 shows three flat disks (of the same radius) that
can rotate about their centers like merry-go-rounds. Each disk con-
sists of the same two materials, one denser than the other (density is
mass per unit volume). In disks 1 and 3, the denser material forms
the outer half of the disk area. In disk 2, it forms the inner half of the
disk area. Forces with identical magnitudes are applied tangentially
to the disk, either at the outer edge or at the interface of the two ma-
terials, as shown. Rank the disks according to (a) the torque about
the disk center, (b) the rotational inertia about the disk center, and
(c) the angular acceleration of the disk, greatest first.
F
Denser
Disk 1
Denser
Disk 3
F F
Lighter
Disk 2
Figure 10-27 Question 10.
11 Figure 10-28ashows a meter stick, half wood and half steel,
that is pivoted at the wood end at O.A force is applied to the steel
end at a. In Fig. 10-28b, the stick is reversed and pivoted at the steel
end at O, and the same force is applied at the wood end at a. Is the
resulting angular acceleration of Fig. 10-28agreater than, less than, or
the same as that of Fig.10-28b?
F
:
R:
M:
1 m
26 kg
(a)
2 m
7 kg
(b)
3 m
3 kg
(c)
Module 10-1 Rotational Variables
•1 A good baseball pitcher can throw a baseball toward home
plate at 85 mi/h with a spin of 1800 rev/min. How many revolutions
does the baseball make on its way to home plate? For simplicity,
assume that the 60 ft path is a straight line.
•2 What is the angular speed of (a) the second hand, (b) the
minute hand, and (c) the hour hand of a smoothly running analog
watch? Answer in radians per second.
••3 When a slice of buttered toast is accidentally pushed
over the edge of a counter, it rotates as it falls. If the distance to the
floor is 76 cm and for rotation less than 1 rev, what are the (a)
smallest and (b) largest angular speeds that cause the toast to hit
and then topple to be butter-side down?
••4 The angular position of a point on a rotating wheel is given
by u2.0 4.0t22.0t3, where uis in radians and tis in seconds.At
t0, what are (a) the point’s angular position and (b) its angular ve-
locity? (c) What is its angular velocity at t4.0 s? (d) Calculate its an-
gular acceleration at t2.0 s. (e) Is its angular acceleration constant?
••5 A diver makes 2.5 revolutions on the way from a 10-m-high
platform to the water. Assuming zero initial vertical velocity, find
the average angular velocity during the dive.
••6 The angular position of a point on the rim of a rotating wheel is
given by u4.0t3.0t2t3, where uis in radians and tis in seconds.
What are the angular velocities at (a) t2.0 s and (b) t4.0 s?
(c) What is the average angular acceleration for the time interval
that begins at t2.0 s and ends at t4.0 s? What are the instanta-
neous angular accelerations at (d) the beginning and (e) the end of
this time interval?
•••7 The wheel in Fig. 10-30 has eight equally spaced spokes and
a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5
rev/s.You want to shoot a 20-cm-long arrow parallel to this axle and
ILW
through the wheel without hitting any
of the spokes. Assume that the arrow
and the spokes are very thin. (a) What
minimum speed must the arrow have?
(b) Does it matter where between the
axle and rim of the wheel you aim? If
so,what is the best location?
•••8 The angular acceleration of a
wheel is a6.0t44.0t2, with ain ra-
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Figure 10-30 Problem 7.
(b)(a)
OO´a´
a
FF
Figure 10-28
Question 11.
Figure 10-29 Question 12.
12 Figure 10-29 shows three disks,
each with a uniform distribution of
mass. The radii Rand masses Mare
indicated. Each disk can rotate
around its central axis (perpendicular
to the disk face and through the cen-
ter). Rank the disks according to their
rotational inertias calculated about
their central axes,greatest first.
dians per second-squared and tin seconds.At time t0, the wheel
has an angular velocity of 2.0 rad/s and an angular position of
1.0 rad.Write expressions for (a) the angular velocity (rad/s) and
(b) the angular position (rad) as functions of time (s).
Module 10-2 Rotation with Constant Angular Acceleration
•9 A drum rotates around its central axis at an angular velocity
of 12.60 rad/s. If the drum then slows at a constant rate of 4.20
rad/s2, (a) how much time does it take and (b) through what angle
does it rotate in coming to rest?
•10 Starting from rest, a disk rotates about its central axis with
constant angular acceleration. In 5.0 s, it rotates 25 rad. During that
time, what are the magnitudes of (a) the angular acceleration and
(b) the average angular velocity? (c) What is the instantaneous an-
gular velocity of the disk at the end of the 5.0 s? (d) With the angu-
lar acceleration unchanged, through what additional angle will the
disk turn during the next 5.0 s?
•11 A disk, initially rotating at 120 rad/s, is slowed down
with a constant angular acceleration of magnitude 4.0 rad/s2. (a) How
much time does the disk take to stop? (b) Through what angle does
the disk rotate during that time?
•12 The angular speed of an automobile engine is increased at a
constant rate from 1200 rev/min to 3000 rev/min in 12 s. (a) What is
a point on Earth’s surface at latitude 40N? (Earth rotates about
that axis.) (b) What is the linear speed vof the point? What are
(c) vand (d) vfor a point at the equator?
••26 The flywheel of a steam engine runs with a constant angular ve-
locity of 150 rev/min.When steam is shut off,the friction of the bearings
and of the air stops the wheel in 2.2 h. (a) What is the constant angular
acceleration, in revolutions per minute-squared, of the wheel during
the slowdown? (b) How many revolutions does the wheel make before
stopping? (c) At the instant the flywheel is turning at 75 rev/min, what
is the tangential component of the linear acceleration of a flywheel par-
ticle that is 50 cm from the axis of rotation? (d) What is the magnitude
of the net linear acceleration of the particle in (c)?
••27 A seed is on a turntable rotating at rev/min, 6.0 cm
from the rotation axis.What are (a) the seed’s acceleration and (b)
the least coefficient of static friction to avoid slippage? (c) If the
turntable had undergone constant angular acceleration from rest
in 0.25 s,what is the least coefficient to avoid slippage?
••28 In Fig. 10-31, wheel Aof radius
rA10 cm is coupled by belt Bto
wheel Cof radius rC25 cm.The an-
gular speed of wheel Ais increased
from rest at a constant rate of
1.6 rad/s2. Find the time needed for
wheel Cto reach an angular speed of
100 rev/min, assuming the belt does
not slip. (Hint: If the belt does not slip, the linear speeds at the two
rims must be equal.)
••29 Figure 10-32 shows an early method of measuring the
speed of light that makes use of a rotating slotted wheel.A beam of
331
3
celeration (in revolutions per minute-squared) will increase the
wheel’s angular speed to 1000 rev/min in 60.0 s? (d) How many
revolutions does the wheel make during that 60.0 s?
•24 A vinyl record is played by rotating the record so that an ap-
proximately circular groove in the vinyl slides under a stylus.
Bumps in the groove run into the stylus, causing it to oscillate.The
equipment converts those oscillations to electrical signals and then
to sound. Suppose that a record turns at the rate of , the
groove being played is at a radius of 10.0 cm, and the bumps in the
groove are uniformly separated by 1.75 mm.At what rate (hits per
second) do the bumps hit the stylus?
••25 (a) What is the angular speed vabout the polar axis of
SSM
331
3 rev/min
288 CHAPTER 10 ROTATION
its angular acceleration in revolutions per minute-squared? (b) How
many revolutions does the engine make during this 12 s interval?
••13 A flywheel turns through 40 rev as it slows from an
angular speed of 1.5 rad/s to a stop. (a) Assuming a constant angu-
lar acceleration, find the time for it to come to rest. (b) What is its
angular acceleration? (c) How much time is required for it to com-
plete the first 20 of the 40 revolutions?
••14 A disk rotates about its central axis starting from rest and
accelerates with constant angular acceleration.At one time it is ro-
tating at 10 rev/s; 60 revolutions later, its angular speed is 15 rev/s.
Calculate (a) the angular acceleration, (b) the time required to
complete the 60 revolutions, (c) the time required to reach the
10 rev/s angular speed, and (d) the number of revolutions from rest
until the time the disk reaches the 10 rev/s angular speed.
••15 Starting from rest, a wheel has constant a= 3.0 rad/s2.
During a certain 4.0 s interval, it turns through 120 rad. How much
time did it take to reach that 4.0 s interval?
••16 A merry-go-round rotates from rest with an angular accel-
eration of 1.50 rad/s2. How long does it take to rotate through
(a) the first 2.00 rev and (b) the next 2.00 rev?
••17 At t0, a flywheel has an angular velocity of 4.7 rad/s, a
constant angular acceleration of 0.25 rad/s2, and a reference line
at u00. (a) Through what maximum angle umax will the reference
line turn in the positive direction? What are the (b) first and
(c) second times the reference line will be at ? At what
(d) negative time and (e) positive time will the reference line be
at 10.5 rad? (f) Graph uversus t, and indicate your answers.
•••18 A pulsar is a rapidly rotating neutron star that emits a radio
beam the way a lighthouse emits a light beam. We receive a radio
pulse for each rotation of the star.The period Tof rotation is found
by measuring the time between pulses.The pulsar in the Crab neb-
ula has a period of rotation of T0.033 s that is increasing at the
rate of 1.26 105s/y. (a) What is the pulsar’s angular acceleration
a? (b) If ais constant, how many years from now will the pulsar
stop rotating? (c) The pulsar originated in a supernova explosion
seen in the year 1054.Assuming constant a, find the initial T.
Module 10-3 Relating the Linear and Angular Variables
•19 What are the magnitudes of (a) the angular velocity, (b) the ra-
dial acceleration, and (c) the tangential acceleration of a spaceship
taking a circular turn of radius 3220 km at a speed of 29 000 km/h?
•20 An object rotates about a fixed axis, and the angular posi-
tion of a reference line on the object is given by u0.40e2t, where
uis in radians and tis in seconds. Consider a point on the object
that is 4.0 cm from the axis of rotation.At t0, what are the mag-
nitudes of the point’s (a) tangential component of acceleration
and (b) radial component of acceleration?
•21 Between 1911 and 1990, the top of the leaning bell
tower at Pisa, Italy, moved toward the south at an average rate of
1.2 mm/y. The tower is 55 m tall. In radians per second, what is the
average angular speed of the tower’s top about its base?
•22 An astronaut is tested in a centrifuge with radius 10 m and
rotating according to u0.30t2. At t5.0 s, what are the magni-
tudes of the (a) angular velocity, (b) linear velocity, (c) tangential
acceleration, and (d) radial acceleration?
•23 A flywheel with a diameter of 1.20 m is rotating
at an angular speed of 200 rev/min. (a) What is the angular speed
of the flywheel in radians per second? (b) What is the linear speed
of a point on the rim of the flywheel? (c) What constant angular ac-
WWWSSM
u
u1
2umax
SSM
ILW
B
C
rA
A
rC
Figure 10-31 Problem 28.
Light
beam
Light
source
Rotating
slotted wheel
Mirror
perpendicular
to light beam
L
Figure 10-32 Problem 29.
289
PROBLEMS
••41 In Fig. 10-37, two particles,
each with mass m0.85 kg, are fas-
tened to each other, and to a rotation
axis at O, by two thin rods, each with
length d5.6 cm and mass M
1.2 kg. The combination rotates
around the rotation axis with the an-
gular speed v0.30 rad/s. Measured
about O, what are the combination’s
(a) rotational inertia and (b) kinetic energy?
••42 The masses and coordinates of four particles are as
follows: 50 g, x2.0 cm, y2.0 cm; 25 g, x0, y4.0 cm; 25 g,
x3.0 cm, y3.0 cm; 30 g, x2.0 cm, y4.0 cm. What
are the rotational inertias of this collection about the (a) x, (b) y,
and (c) zaxes? (d) Suppose that we symbolize the answers to (a)
and (b) as Aand B, respectively. Then what is the answer to (c)
in terms of Aand B?
L
O
light passes through one of the slots at the outside edge of the
wheel, travels to a distant mirror, and returns to the wheel just in
time to pass through the next slot in the wheel. One such slotted
wheel has a radius of 5.0 cm and 500 slots around its edge.
Measurements taken when the mirror is L500 m from the
wheel indicate a speed of light of 3.0 105km/s. (a) What is the
(constant) angular speed of the wheel? (b) What is the linear
speed of a point on the edge of the wheel?
••30 A gyroscope flywheel of radius 2.83 cm is accelerated from
rest at 14.2 rad/s2until its angular speed is 2760 rev/min. (a) What is
the tangential acceleration of a point on the rim of the flywheel during
this spin-up process? (b) What is the radial acceleration of this point
when the flywheel is spinning at full speed? (c) Through what distance
does a point on the rim move during the spin-up?
••31 A disk, with a radius of 0.25 m, is to be rotated like a merry-
go-round through 800 rad, starting from rest, gaining angular speed
at the constant rate a1through the first 400 rad and then losing an-
gular speed at the constant rate a1until it is again at rest.The mag-
nitude of the centripetal acceleration of any portion of the disk is
not to exceed 400 m/s2. (a) What is the least time required for the ro-
tation? (b) What is the corresponding value of a1?
••32 A car starts from rest and moves around a circular track of
radius 30.0 m. Its speed increases at the constant rate of 0.500 m/s2.
(a) What is the magnitude of its net linear acceleration 15.0 s later?
(b) What angle does this net acceleration vector make with the
car’s velocity at this time?
Module 10-4 Kinetic Energy of Rotation
•33 Calculate the rota-
tional inertia of a wheel that has
a kinetic energy of 24 400 J when
rotating at 602 rev/min.
•34 Figure 10-33 gives angu-
lar speed versus time for a thin
rod that rotates around one
end. The scale on the vaxis is
set by (a) What
is the magnitude of the rod’s an-
gular acceleration? (b) At t
4.0 s, the rod has a rotational ki-
netic energy of 1.60 J.What is its kinetic energy at t0?
Module 10-5 Calculating the Rotational Inertia
•35 Two uniform solid cylinders, each rotating about its cen-
tral (longitudinal) axis at 235 rad/s, have the same mass of 1.25 kg but
differ in radius.What is the rotational kinetic energy of (a) the smaller
cylinder, of radius 0.25 m,and (b) the larger cylinder,of radius 0.75 m?
•36 Figure 10-34ashows a disk that can rotate about an axis at
SSM
vs6.0 rad/s.
SSM
a radial distance hfrom the center of the disk. Figure 10-34bgives
the rotational inertia Iof the disk about the axis as a function of that
distance h, from the center out to the edge of the disk. The scale on
the Iaxis is set by and What is
the mass of the disk?
•37 Calculate the rotational inertia of a meter stick, with
mass 0.56 kg, about an axis perpendicular to the stick and located
at the 20 cm mark. (Treat the stick as a thin rod.)
•38 Figure 10-35 shows three
0.0100 kg particles that have been
glued to a rod of length L6.00 cm
and negligible mass. The assembly
can rotate around a perpendicular
axis through point Oat the left end.
If we remove one particle (that is,
33% of the mass), by what percent-
age does the rotational inertia of the assembly around the rotation
axis decrease when that removed particle is (a) the innermost one
and (b) the outermost one?
••39 Trucks can be run on energy stored in a rotating flywheel,
with an electric motor getting the flywheel up to its top speed of
200prad/s. Suppose that one such flywheel is a solid, uniform
cylinder with a mass of 500 kg and a radius of 1.0 m. (a) What is the
kinetic energy of the flywheel after charging? (b) If the truck uses
an average power of 8.0 kW, for how many minutes can it operate
between chargings?
••40 Figure 10-36 shows an arrangement of 15 identical disks that
have been glued together in a rod-like shape of length L1.0000 m
and (total) mass M100.0 mg.The disks are uniform, and the disk
arrangement can rotate about a perpendicular axis through its cen-
tral disk at point O. (a) What is the rotational inertia of the
arrangement about that axis? (b) If we approximated the arrange-
ment as being a uniform rod of mass Mand length L, what percent-
age error would we make in using the formula in Table 10-2eto cal-
culate the rotational inertia?
SSM
IB0.150 kgm2.IA0.050 kgm2
0
0123456
t (s)
(rad/s)
s
ω
Figure 10-33 Problem 34.
I (kg m2)
I
B
IA0 0.1
h(m)
0.2
(b)(a)
Axis
h
Figure 10-34 Problem 36.
Figure 10-36 Problem 40.
Figure 10-35 Problems
38 and 62.
Axis
L
m
O
d d d
m m
Rotation axis
m
d
m
d
M
M
O
ω
Figure 10-37 Problem 41.
290 CHAPTER 10 ROTATION
••53 Figure 10-43 shows a uniform
disk that can rotate around its center like a
merry-go-round. The disk has a radius of
2.00 cm and a mass of 20.0 grams and is ini-
tially at rest. Starting at time t0, two
forces are to be applied tangentially to the
rim as indicated, so that at time t1.25 s
the disk has an angular velocity of 250
rad/s counterclockwise. Force
has a magnitude of 0.100 N. What
is magnitude F2?
••54 In a judo foot-sweep
move, you sweep your opponent’s
left foot out from under him while
pulling on his gi (uniform) toward
that side. As a result, your oppo-
nent rotates around his right foot
and onto the mat. Figure 10-44
shows a simplified diagram of
your opponent as you face him,
with his left foot swept out. The
rotational axis is through point O.
The gravitational force on him
effectively acts at his center of
mass, which is a horizontal dis-
tance d28 cm from point O.His
mass is 70 kg, and his rotational in-
ertia about point Ois 65 kgm2.What is the magnitude of his initial
angular acceleration about point Oif your pull on his gi is (a) neg-
ligible and (b) horizontal with a magnitude of 300 N and applied at
height h1.4 m?
••55 In Fig. 10-45a, an irregularly shaped plastic plate with
uniform thickness and density (mass per unit volume) is to be
rotated around an axle that is perpendicular to the plate face
and through point O. The rotational inertia of the plate about
F
:
a
F
:
g
F
:
1
••43 The uniform solid
block in Fig. 10-38 has mass 0.172 kg
and edge lengths a3.5 cm, b8.4
cm, and c1.4 cm. Calculate its rota-
tional inertia about an axis through
one corner and perpendicular to the
large faces.
••44 Four identical particles of
mass 0.50 kg each are placed at the
vertices of a 2.0 m 2.0 m square
and held there by four massless rods, which form the sides of the
square. What is the rotational inertia of this rigid body about an
axis that (a) passes through the midpoints of opposite sides and
lies in the plane of the square, (b) passes through the midpoint of
one of the sides and is perpendicular to the plane of the square,
and (c) lies in the plane of the square and passes through two di-
agonally opposite particles?
WWWSSM rest, block 2 falls 75.0 cm in 5.00 s without the cord slipping on
the pulley. (a) What is the magnitude of the acceleration of the
blocks? What are (b) tension and (c) tension ? (d) What is
the magnitude of the pulley’s angular acceleration? (e) What is
its rotational inertia?
••52 In Fig. 10-42, a cylinder having a mass of 2.0 kg can rotate
about its central axis through point O. Forces are applied as shown:
F16.0 N, F24.0 N, F32.0 N, and F45.0 N. Also, r5.0 cm
and R12 cm. Find the (a) magnitude and (b) direction of the an-
gular acceleration of the cylinder. (During the rotation, the forces
maintain their same angles relative to the cylinder.)
T
1
T
2
b
a
c
Rotat
i
on
axis
Figure 10-38 Problem 43.
F2
F1
Figure 10-43
Problem 53.
Figure 10-42 Problem 52.
F1
F4R
r
O
Rotation
axis
F2
F3
Figure 10-39 Problem 45.
O
r1
θ
1
F1
r2
θ
2
F2
Module 10-6 Torque
•45 The body in
Fig. 10-39 is pivoted at O, and
two forces act on it as shown. If
r11.30 m, r22.15 m, F1
4.20 N, F24.90 N, u175.0,
and u260.0, what is the net
torque about the pivot?
•46 The body in Fig. 10-40 is
pivoted at O. Three forces act
on it: FA10 N at point A, 8.0
m from O;FB16 N at B, 4.0
m from O; and FC19 N at C,
3.0 m from O. What is the net
torque about O?
•47 A small ball of mass
0.75 kg is attached to one end
of a 1.25-m-long massless rod,
SSM
ILWSSM
and the other end of the rod is hung from a pivot.When the resulting
pendulum is 30from the vertical, what is the magnitude of the gravi-
tational torque calculated about the pivot?
•48 The length of a bicycle pedal arm is 0.152 m, and a down-
ward force of 111 N is applied to the pedal by the rider.What is the
magnitude of the torque about the pedal arm’s pivot when the arm
is at angle (a) 30, (b) 90, and (c) 180with the vertical?
Module 10-7 Newton’s Second Law for Rotation
•49 During the launch from a board, a diver’s angular
speed about her center of mass changes from zero to 6.20 rad/s in 220
ms. Her rotational inertia about her center of mass is 12.0 kg m2.
During the launch, what are the magnitudes of (a) her average angu-
lar acceleration and (b) the average external
torque on her from the board?
•50 If a 32.0 N m torque on a wheel
causes angular acceleration 25.0 rad/s2,
what is the wheel’s rotational inertia?
••51 In Fig. 10-41, block 1 has mass
, block 2 has mass ,
and the pulley, which is mounted on a hor-
izontal axle with negligible friction, has
radius . When released fromR5.00 cm
m2500 gm1460 g
ILWSSM
Figure 10-44 Problem 54.
O
d
com
Fg
Fa
h
FCC
FB
B
160°
90° O
FA
A
135°
Figure 10-40 Problem 46.
m1
T1T2
m2
R
Figure 10-41
Problems 51 and 83.
291
PROBLEMS
that axle is measured with the following
method. A circular disk of mass 0.500 kg and
radius 2.00 cm is glued to the plate, with its
center aligned with point O(Fig. 10-45b). A
string is wrapped around the edge of the disk
the way a string is wrapped around a top.
Then the string is pulled for 5.00 s. As a re-
sult, the disk and plate are rotated by a con-
stant force of 0.400 N that is applied by the
string tangentially to the edge of the disk.
The resulting angular speed is 114 rad/s.
What is the rotational inertia of the plate
about the axle?
••56 Figure 10-46 shows
particles 1 and 2, each of mass
m, fixed to the ends of a rigid
massless rod of length L1
L2, with L120 cm and L2
80 cm. The rod is held hori-
zontally on the fulcrum and then released. What are the magni-
tudes of the initial accelerations of (a) particle 1 and (b) particle 2?
•••57 A pulley, with a rotational inertia of 1.0 103kgm2about
its axle and a radius of 10 cm, is acted on by a force applied tangentially
at its rim.The force magnitude varies in time as F0.50t0.30t2, with
Fin newtons and tin seconds.The pulley is initially at rest.At t3.0 s
what are its (a) angular acceleration and (b) angular speed?
Module 10-8 Work and Rotational Kinetic Energy
•58 (a) If R12 cm, M400 g, and m50 g in Fig. 10-19, find
the speed of the block after it has descended 50 cm starting from
rest. Solve the problem using energy conservation principles.
(b) Repeat (a) with R5.0 cm.
•59 An automobile crankshaft transfers energy from the engine
to the axle at the rate of 100 hp (74.6 kW) when rotating at a
speed of 1800 rev/min. What torque (in newton-meters) does the
crankshaft deliver?
•60 A thin rod of length 0.75 m and mass 0.42 kg is suspended
freely from one end. It is pulled to one side and then allowed to swing
like a pendulum, passing through its lowest position with angular
speed 4.0 rad/s. Neglecting friction and air resistance, find (a) the
rod’s kinetic energy at its lowest position and (b) how far above that
position the center of mass rises.
•61 A 32.0 kg wheel, essentially a thin hoop with radius 1.20 m, is
rotating at 280 rev/min. It must be brought to a stop in 15.0 s.
(a) How much work must be done to stop it? (b) What is the
required average power?
••62 In Fig. 10-35, three 0.0100 kg particles have been glued to a
rod of length L6.00 cm and negligible mass and can rotate
around a perpendicular axis through point Oat one end. How
much work is required to change the rotational rate (a) from 0 to
20.0 rad/s, (b) from 20.0 rad/s to 40.0 rad/s, and (c) from 40.0 rad/s to
60.0 rad/s? (d) What is the slope of a plot of the assembly’s kinetic
energy (in joules) versus the square of its rotation rate (in radians-
squared per second-squared)?
••63 A meter stick is held vertically with one end on
the floor and is then allowed to fall. Find the speed of the other end
just before it hits the floor, assuming that the end on the floor does
not slip. (Hint: Consider the stick to be a thin rod and use the con-
servation of energy principle.)
ILWSSM
••64 A uniform cylinder of radius 10 cm and mass 20 kg is
mounted so as to rotate freely about a horizontal axis that is paral-
lel to and 5.0 cm from the central longitudinal axis of the cylinder.
(a) What is the rotational inertia of the cylinder about the axis of
rotation? (b) If the cylinder is released from rest with its central
longitudinal axis at the same height as the axis about which the
cylinder rotates, what is the angular speed of the cylinder as it
passes through its lowest position?
•••65 A tall, cylindrical chimney falls over when its base
is ruptured.Treat the chimney as a thin rod of length 55.0 m. At the
instant it makes an angle of 35.0with the vertical as it falls, what
are (a) the radial acceleration of the top, and (b) the tangential ac-
celeration of the top. (Hint: Use energy considerations, not a torque.)
(c) At what angle uis the tangential acceleration equal to g?
•••66 A uniform spherical shell of mass M4.5 kg and radius
(b)
(a)
O
Axle
Plate
Disk
String
Figure 10-45
Problem 55.
1 2
L1L2
Figure 10-46 Problem 56.
Figure 10-47 Problem 66.
R8.5 cm can rotate about a vertical axis on frictionless bearings
(Fig. 10-47). A massless cord passes around the equator of the shell,
over a pulley of rotational inertia I3.0 103kgm2and radius
r5.0 cm, and is attached to a small object of mass m0.60 kg.
There is no friction on the pulley’s axle; the cord does not slip on
the pulley.What is the speed of the object when it has fallen 82 cm
after being released from rest? Use energy considerations.
M, R
I, r
m
•••67 Figure 10-48 shows a rigid as-
sembly of a thin hoop (of mass mand ra-
dius R0.150 m) and a thin radial rod
(of mass mand length L2.00R). The
assembly is upright, but if we give it a
slight nudge, it will rotate around a hori-
zontal axis in the plane of the rod and
hoop, through the lower end of the rod.
Assuming that the energy given to the
assembly in such a nudge is negligible,
what would be the assembly’s angular speed about the rotation axis
when it passes through the upside-down (inverted) orientation?
Additional Problems
68 Two uniform solid spheres have the same mass of 1.65 kg, but
one has a radius of 0.226 m and the other has a radius of 0.854 m.
Each can rotate about an axis through its center. (a) What is the
magnitude tof the torque required to bring the smaller sphere
from rest to an angular speed of 317 rad/s in 15.5 s? (b) What is the
magnitude Fof the force that must be
applied tangentially at the sphere’s
equator to give that torque? What are
the corresponding values of (c) tand
(d) Ffor the larger sphere?
69 In Fig. 10-49, a small disk of radius
r2.00 cm has been glued to the edge of
a larger disk of radius R4.00 cm so that
Figure 10-48 Problem 67.
Rotation
a
xi
s
Hoop
Rod
O
Figure 10-49 Problem 69.
292 CHAPTER 10 ROTATION
the disks lie in the same plane. The disks can be rotated around a per-
pendicular axis through point Oat the center of the larger disk. The
disks both have a uniform density (mass per unit volume) of 1.40
103kg/m3and a uniform thickness of 5.00 mm. What is the rota-
tional inertia of the two-disk assembly about the rotation axis
through O?
70 A wheel, starting from rest, rotates with a constant angular
acceleration of 2.00 rad/s2. During a certain 3.00 s interval, it turns
through 90.0 rad. (a) What is the angular velocity of the wheel at
the start of the 3.00 s interval? (b) How long has the wheel been
turning before the start of the 3.00 s interval?
71 In Fig. 10-50, two 6.20 kg
blocks are connected by a massless
string over a pulley of radius 2.40 cm
and rotational inertia 7.40 104
kgm2. The string does not slip on
the pulley; it is not known whether
there is friction between the table and
the sliding block; the pulley’s axis is
frictionless. When this system is re-
leased from rest, the pulley turns through 0.130 rad in 91.0 ms and the
acceleration of the blocks is constant. What are (a) the magnitude of
the pulley’s angular acceleration, (b) the magnitude of either block’s
acceleration,(c) string tension T1, and (d) string tension T2?
72 Attached to each end of a thin steel rod of length 1.20 m and
mass 6.40 kg is a small ball of mass 1.06 kg. The rod is constrained
to rotate in a horizontal plane about a vertical axis through its mid-
point.At a certain instant, it is rotating at 39.0 rev/s. Because of fric-
tion, it slows to a stop in 32.0 s.Assuming a constant retarding torque
due to friction, compute (a) the angular acceleration, (b) the retard-
ing torque, (c) the total energy transferred from mechanical energy
to thermal energy by friction, and (d) the number of revolutions ro-
tated during the 32.0 s. (e) Now suppose that the retarding torque is
known not to be constant. If any of the quantities (a), (b),(c), and (d)
can still be computed without additional information, give its value.
73 A uniform helicopter rotor blade is 7.80 m long, has a mass of
110 kg, and is attached to the rotor axle by a single bolt. (a) What is
the magnitude of the force on the bolt from the axle when the ro-
tor is turning at 320 rev/min? (Hint: For this calculation the blade
can be considered to be a point mass at its center of mass. Why?)
(b) Calculate the torque that must be applied to the rotor to bring
it to full speed from rest in 6.70 s. Ignore air resistance. (The blade
cannot be considered to be a point mass for this calculation. Why
not? Assume the mass distribution of a uniform thin rod.) (c) How
much work does the torque do on the blade in order for the blade
to reach a speed of 320 rev/min?
74 Racing disks. Figure 10-51 shows
two disks that can rotate about their
centers like a merry-go-round. At
time t0, the reference lines of the
two disks have the same orientation.
Disk Ais already rotating, with a con-
stant angular velocity of 9.5 rad/s.
Disk Bhas been stationary but now begins to rotate at a constant
angular acceleration of 2.2 rad/s2. (a) At what time twill the refer-
ence lines of the two disks momentarily have the same angular dis-
placement u? (b) Will that time tbe the first time since t0 that
the reference lines are momentarily aligned?
75 A high-wire walker always attempts to keep his center of
mass over the wire (or rope). He normally carries a long, heavy pole
SSM
to help: If he leans, say, to his right (his com moves to the right) and is
in danger of rotating around the wire, he moves the pole to his left
(its com moves to the left) to slow the rotation and allow himself
time to adjust his balance. Assume that the walker has a mass of
70.0 kg and a rotational inertia of about the wire.What is
the magnitude of his angular acceleration about the wire if his com is
5.0 cm to the right of the wire and (a) he carries no pole and (b) the
14.0 kg pole he carries has its com 10 cm to the left of the wire?
76 Starting from rest at t0, a wheel undergoes a constant an-
gular acceleration. When t2.0 s, the angular velocity of the
wheel is 5.0 rad/s.The acceleration continues until t20 s, when it
abruptly ceases. Through what angle does the wheel rotate in the
interval t0 to t40 s?
77 A record turntable rotating at slows down
and stops in 30 s after the motor is turned off. (a) Find its (con-
stant) angular acceleration in revolutions per minute-squared.
(b) How many revolutions does it make in this time?
78 A rigid body is made of three
identical thin rods, each with length
L0.600 m, fastened together in the
form of a letter H(Fig. 10-52). The
body is free to rotate about a hori-
zontal axis that runs along the length
of one of the legs of the H. The body
is allowed to fall from rest from a position in which the plane of the
His horizontal. What is the angular speed of the body when the
plane of the His vertical?
79 (a) Show that the rotational inertia of a solid cylinder of
mass Mand radius Rabout its central axis is equal to the rotational
inertia of a thin hoop of mass Mand radius about its central
axis. (b) Show that the rotational inertia Iof any given body of
mass Mabout any given axis is equal to the rotational inertia of an
equivalent hoop about that axis, if the hoop has the same mass M
and a radius kgiven by
The radius kof the equivalent hoop is called the radius of gyration
of the given body.
80 A disk rotates at constant angular acceleration, from angular
position u110.0 rad to angular position u270.0 rad in 6.00 s. Its
angular velocity at u2is 15.0 rad/s. (a) What was its angular velocity
at u1? (b) What is the angular acceleration? (c) At what angular
position was the disk initially at rest? (d) Graph uversus time tand
angular speed vversus tfor the disk, from the beginning of the
motion (let t0 then).
81 The thin uniform rod in Fig. 10-53 has
length 2.0 m and can pivot about a horizontal,
frictionless pin through one end. It is released
from rest at angle u40above the horizontal.
Use the principle of conservation of energy to
determine the angular speed of the rod as it
passes through the horizontal position.
82 George Washington Gale Ferris, Jr., a
civil engineering graduate from Rensselaer Polytechnic Institute,
built the original Ferris wheel for the 1893 World’s Columbian
Exposition in Chicago. The wheel, an astounding engineering con-
struction at the time, carried 36 wooden cars, each holding up to 60
passengers, around a circle 76 m in diameter.The cars were loaded 6
at a time, and once all 36 cars were full, the wheel made a complete
kAI
M.
R/22
SSM
331
3 rev/min
SSM
15.0 kgm2
T2
T1
Figure 10-51 Problem 74.
Di
s
kADi
s
kB
L
L
L
Figure 10-52 Problem 78.
Figure 10-50 Problem 71.
θ
Pin
Figure 10-53
Problem 81.
the rings are given in the following table. A tangential force of
magnitude 12.0 N is applied to the outer edge of the outer ring for
0.300 s.What is the change in the angular speed of the construction
during the time interval?
Ring Mass (kg) Inner Radius (m) Outer Radius (m)
1 0.120 0.0160 0.0450
2 0.240 0.0900 0.1400
87 In Fig. 10-55, a wheel of ra-
dius 0.20 m is mounted on a friction-
less horizontal axle. A massless cord
is wrapped around the wheel and at-
tached to a 2.0 kg box that slides on
a frictionless surface inclined at an-
gle u20with the horizontal. The
box accelerates down the surface at 2.0 m/s2. What is the rota-
tional inertia of the wheel about the axle?
88 A thin spherical shell has a radius of 1.90 m.An applied torque
of 960 Nm gives the shell an angular acceleration of 6.20 rad/s2
about an axis through the center of the shell. What are (a) the rota-
tional inertia of the shell about that axis and (b) the mass of the shell?
89 A bicyclist of mass 70 kg puts all his mass on each downward-
moving pedal as he pedals up a steep road. Take the diameter of
293
PROBLEMS
rotation at constant angular speed in about 2 min. Estimate the
amount of work that was required of the machinery to rotate the
passengers alone.
83 In Fig.10-41, two blocks,of mass m1400 g and m2600 g,are
connected by a massless cord that is wrapped around a uniform disk
of mass M500 g and radius R12.0 cm.The disk can rotate with-
out friction about a fixed horizontal axis through its center; the cord
cannot slip on the disk.The system is released from rest. Find (a) the
magnitude of the acceleration of the blocks, (b) the tension T1in the
cord at the left,and (c) the tension T2in the cord at the right.
84 At 714 A.M. on June 30, 1908, a huge explosion
the circle in which the pedals rotate to be 0.40 m, and determine
the magnitude of the maximum torque he exerts about the rota-
tion axis of the pedals.
90 The flywheel of an engine is rotating at 25.0 rad/s. When the
engine is turned off, the flywheel slows at a constant rate and stops
in 20.0 s. Calculate (a) the angular acceleration of the flywheel,
(b) the angle through which the flywheel rotates in stopping, and
(c) the number of revolutions made by the flywheel in stopping.
91 In Fig. 10-19a, a wheel of radius 0.20 m is mounted on a fric-
tionless horizontal axis. The rotational inertia of the wheel about the
axis is 0.40 kg m2. A massless cord wrapped around the wheel’s cir-
cumference is attached to a 6.0 kg box. The system is released from
rest.When the box has a kinetic energy of 6.0 J, what are (a) the wheel’s
rotational kinetic energy and (b) the distance the box has fallen?
92 Our Sun is 2.3 104ly (light-years) from the center of our
Milky Way galaxy and is moving in a circle around that center at a
speed of 250 km/s. (a) How long does it take the Sun to make one
revolution about the galactic center? (b) How many revolutions has
the Sun completed since it was formed about 4.5 109years ago?
93 A wheel of radius 0.20 m
is mounted on a frictionless horizon-
tal axis. The rotational inertia of the
wheel about the axis is 0.050 kgm2.
A massless cord wrapped around
the wheel is attached to a 2.0 kg
block that slides on a horizontal frictionless surface. If a horizontal
force of magnitude P3.0 N is applied to the block as shown in
Fig. 10-56, what is the magnitude of the angular acceleration of the
wheel? Assume the cord does not slip on the wheel.
94 If an airplane propeller rotates at 2000 rev/min while the air-
plane flies at a speed of 480 km/h relative to the ground, what is the
linear speed of a point on the tip of the propeller, at radius 1.5 m, as
seen by (a) the pilot and (b) an observer on the ground? The plane’s
velocity is parallel to the propeller’s axis of rotation.
95 The rigid body shown in
Fig. 10-57 consists of three particles
connected by massless rods. It is to be
rotated about an axis perpendicular
to its plane through point P. If M
0.40 kg, a30 cm, and b50 cm,
how much work is required to take
the body from rest to an angular
speed of 5.0 rad/s?
96 Beverage engineering. The pull
tab was a major advance in the engi-
neering design of beverage contain-
ers.The tab pivots on a central bolt in the can’s top.When you pull
upward on one end of the tab, the other end presses downward on
a portion of the can’s top that has been scored. If you pull upward
with a 10 N force, what force magnitude acts on the scored section?
(You will need to examine a can with a pull tab.)
97 Figure 10-58 shows a propeller blade that rotates at
2000 rev/min about a perpendicular axis at point B. Point Ais at
the outer tip of the blade, at radial distance
1.50 m. (a) What is the difference in the
magnitudes aof the centripetal acceleration
of point Aand of a point at radial distance
0.150 m? (b) Find the slope of a plot of a
versus radial distance along the blade.
SSM
SSM
86 Figure 10-54 shows a flat construction of
two circular rings that have a common center and
are held together by three rods of negligible mass.
The construction, which is initially at rest, can
rotate around the common center (like a merry-
go-round), where another rod of negligible mass
lies. The mass, inner radius, and outer radius of
occurred above remote central Siberia, at latitude 61N and lon-
gitude 102E; the fireball thus created was the brightest flash
seen by anyone before nuclear weapons. The Tunguska Event,
which according to one chance witness “covered an enormous part
of the sky, was probably the explosion of a stony asteroid about 140
m wide. (a) Considering only Earth’s rotation, determine how much
later the asteroid would have had to arrive to put the explosion
above Helsinki at longitude 25E. This would have obliterated the
city. (b) If the asteroid had, instead, been a metallic asteroid, it could
have reached Earth’s surface. How much later would such an
asteroid have had to arrive to put the impact in the Atlantic Ocean
at longitude 20W? (The resulting tsunamis would have wiped out
coastal civilization on both sides of the Atlantic.)
85 A golf ball is launched at an angle of 20to the horizontal,
with a speed of 60 m/s and a rotation rate of 90 rad/s.Neglecting air
drag, determine the number of revolutions the ball makes by the
time it reaches maximum height.
Figure 10-54
Problem 86.
Figure 10-56 Problem 93.
P
P
b b
M
2M2M
a a
Figure 10-57 Problem 95.
Figure 10-55 Problem 87.
θ
B
A
Figure 10-58
Problem 97.
294 CHAPTER 10 ROTATION
98 A yo-yo-shaped device
mounted on a horizontal fric-
tionless axis is used to lift a 30 kg
box as shown in Fig. 10-59. The
outer radius Rof the device is
0.50 m, and the radius rof the
hub is 0.20 m. When a constant
horizontal force of magni-
tude 140 N is applied to a rope
wrapped around the outside of
the device, the box, which is sus-
pended from a rope wrapped
around the hub, has an upward
acceleration of magnitude 0.80
m/s2.What is the rotational iner-
tia of the device about its axis of rotation?
99 A small ball with mass 1.30 kg is mounted on one end of a rod
0.780 m long and of negligible mass. The system rotates in a horizon-
tal circle about the other end of the rod at 5010 rev/min. (a) Calculate
the rotational inertia of the system about the axis of rotation. (b)
There is an air drag of 2.30 102N on the ball, directed opposite its
motion. What torque must be applied to the system to keep it rotat-
ing at constant speed?
100 Two thin rods (each of mass 0.20
kg) are joined together to form a rigid
body as shown in Fig. 10-60. One of the
rods has length L10.40 m, and the
other has length L20.50 m. What is
the rotational inertia of this rigid body
about (a) an axis that is perpendicular
to the plane of the paper and passes
through the center of the shorter rod
and (b) an axis that is perpendicular to
the plane of the paper and passes
through the center of the longer rod?
101 In Fig. 10-61, four pul-
leys are connected by two
belts. Pulley A(radius 15 cm)
is the drive pulley, and it ro-
tates at 10 rad/s. Pulley B(ra-
dius 10 cm) is connected by
belt 1 to pulley A. Pulley B
(radius 5 cm) is concentric with
pulley Band is rigidly attached
to it. Pulley C(radius 25 cm) is
connected by belt 2 to pulley B.
Calculate (a) the linear speed of
a point on belt 1, (b) the angular
F
:
app
and three connecting rods, with and
. The balls may be treated as particles, and the connecting
rods have negligible mass. Determine the rotational kinetic energy
of the object if it has an angular speed of 1.2 rad/s about (a) an axis
that passes through point Pand is perpendicular to the plane of the
figure and (b) an axis that passes through point P, is perpendicular
to the rod of length 2L, and lies in the plane of the figure.
103 In Fig. 10-63, a thin uniform rod
(mass 3.0 kg, length 4.0 m) rotates
freely about a horizontal axis Athat is
perpendicular to the rod and passes
through a point at distance d1.0 m
from the end of the rod. The kinetic
energy of the rod as it passes through
the vertical position is 20 J. (a) What is
the rotational inertia of the rod about
axis A? (b) What is the (linear) speed
of the end Bof the rod as the rod
passes through the vertical position?
(c) At what angle uwill the rod mo-
mentarily stop in its upward swing?
104 Four particles, each of mass,
0.20 kg, are placed at the vertices of a
square with sides of length 0.50 m.The
particles are connected by rods of neg-
ligible mass. This rigid body can rotate
in a vertical plane about a horizontal
axis Athat passes through one of the
particles. The body is released from
rest with rod AB horizontal (Fig. 10-64).
(a) What is the rotational inertia of the
body about axis A? (b) What is the an-
gular speed of the body about axis A
when rod AB swings through the verti-
cal position?
105 Cheetahs running at top speed have been reported at an as-
tounding 114 km/h (about 71 mi/h) by observers driving alongside
the animals. Imagine trying to measure a cheetah’s speed by keeping
your vehicle abreast of the animal while also glancing at your
speedometer, which is registering 114 km/h. You keep the vehicle a
constant 8.0 m from the cheetah, but the noise of the vehicle causes
the cheetah to continuously veer away from you along a circular
path of radius 92 m. Thus, you travel along a circular path of radius
100 m. (a) What is the angular speed of you and the cheetah around
the circular paths? (b) What is the linear speed of the cheetah along
its path? (If you did not account for the circular motion, you would
conclude erroneously that the cheetah’s speed is 114 km/h, and that
type of error was apparently made in the published reports.)
106 A point on the rim of a 0.75-m-diameter grinding wheel
changes speed at a constant rate from 12 m/s to 25 m/s in 6.2 s.
What is the average angular acceleration of the wheel?
107 A pulley wheel that is 8.0 cm in diameter has a 5.6-m-long
cord wrapped around its periphery. Starting from rest, the wheel is
given a constant angular acceleration of 1.5 rad/s2. (a) Through
what angle must the wheel turn for the cord to unwind com-
pletely? (b) How long will this take?
108 A vinyl record on a turntable rotates at 33 rev/min.
(a) What is its angular speed in radians per second? What is the
linear speed of a point on the record (b) 15 cm and (c) 7.4 cm from
the turntable axis?
1
3
30
M1.6 kg, L0.60 m,
R
r
Rope
Yo-yo-shaped
device
Rigid mount
Hub
Fapp
Figure 10-59 Problem 98.
A
BBelt 1
Drive
pulley
Belt 2
B'
C
Figure 10-61 Problem 101.
Figure 10-60 Problem 100.
L1
L2
1
__
2L1
1
__
2
θ
θ
2M
2M
M
2L
L
L
P
Figure 10-62
Problem 102.
Figure 10-63 Problem 103.
θ
A
B
d
Figure 10-64 Problem 104.
A B
Rotation
axis
speed of pulley B, (c) the angular speed of pulley B, (d) the linear
speed of a point on belt 2, and (e) the angular speed of pulley C.(Hint:
If the belt between two pulleys does not slip, the linear speeds at the
rims of the two pulleys must be equal.)
102 The rigid object shown in Fig. 10-62 consists of three balls
CHAPTER 11
Rolling, Torque, and Angular Momentum
295295
11-1 ROLLING AS TRANSLATION AND ROTATION COMBINED
After reading this module, you should be able to . . .
11.01 Identify that smooth rolling can be considered as a
combination of pure translation and pure rotation.
11.02 Apply the relationship between the center-of-mass
speed and the angular speed of a body in smooth rolling.
For a wheel of radius Rrolling smoothly,
vcom vR,
where vcom is the linear speed of the wheels center of mass
and vis the angular speed of the wheel about its center.
The wheel may also be viewed as rotating instantaneously
about the point Pof the “road” that is in contact with the
wheel. The angular speed of the wheel about this point
is the same as the angular speed of the wheel about
its center.
Learning Objectives
Key Ideas
What Is Physics?
As we discussed in Chapter 10, physics includes the study of rotation.Arguably,
the most important application of that physics is in the rolling motion of wheels
and wheel-like objects. This applied physics has long been used. For example,
when the prehistoric people of Easter Island moved their gigantic stone statues
from the quarry and across the island, they dragged them over logs acting as
rollers. Much later, when settlers moved westward across America in the 1800s,
they rolled their possessions first by wagon and then later by train.Today, like it
or not, the world is filled with cars, trucks, motorcycles, bicycles, and other
rolling vehicles.
The physics and engineering of rolling have been around for so long that
you might think no fresh ideas remain to be developed. However, skateboards
and inline skates were invented and engineered fairly recently, to become huge
financial successes. Street luge is now catching on, and the self-righting Segway
(Fig. 11-1) may change the way people move around in large cities. Applying the
physics of rolling can still lead to surprises and rewards. Our starting point in
exploring that physics is to simplify rolling motion.
Rolling as Translation and Rotation Combined
Here we consider only objects that roll smoothly along a surface; that is, the
objects roll without slipping or bouncing on the surface. Figure 11-2 shows how
complicated smooth rolling motion can be: Although the center of the object
moves in a straight line parallel to the surface, a point on the rim certainly
does not. However, we can study this motion by treating it as a combination
of translation of the center of mass and rotation of the rest of the object around
that center.
Figure 11-1 The self-righting Segway Human
Transporter.
Justin Sullivan/Getty Images, Inc.
To see how we do this, pretend you are standing on a sidewalk watching the
bicycle wheel of Fig. 11-3 as it rolls along a street.As shown, you see the center of
mass Oof the wheel move forward at constant speed vcom. The point Pon the
street where the wheel makes contact with the street surface also moves forward
at speed vcom, so that Palways remains directly below O.
During a time interval t, you see both Oand Pmove forward by a distance s.
The bicycle rider sees the wheel rotate through an angle uabout the center of the
wheel, with the point of the wheel that was touching the street at the beginning
of tmoving through arc length s. Equation 10-17 relates the arc length sto the
rotation angle u:
suR, (11-1)
where Ris the radius of the wheel. The linear speed vcom of the center of the
wheel (the center of mass of this uniform wheel) is ds/dt. The angular speed vof
the wheel about its center is du/dt. Thus, differentiating Eq. 11-1 with respect to
time (with Rheld constant) gives us
vcom vR(smooth rolling motion). (11-2)
A Combination. Figure 11-4 shows that the rolling motion of a wheel is a
combination of purely translational and purely rotational motions. Figure 11-4a
shows the purely rotational motion (as if the rotation axis through the center
were stationary): Every point on the wheel rotates about the center with angular
speed v. (This is the type of motion we considered in Chapter 10.) Every point on
the outside edge of the wheel has linear speed vcom given by Eq. 11-2. Figure 11-4b
shows the purely translational motion (as if the wheel did not rotate at all): Every
point on the wheel moves to the right with speed vcom.
The combination of Figs. 11-4aand 11-4byields the actual rolling motion
of the wheel, Fig. 11-4c. Note that in this combination of motions, the portion
of the wheel at the bottom (at point P) is stationary and the portion at the top
296 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Figure 11-2 A time-exposure photo-
graph of a rolling disk. Small lights
have been attached to the disk, one
at its center and one at its edge.
The latter traces out a curve called
acycloid.
Richard Megna/Fundamental Photographs
P
O
vcom
T
v = 2vcom
v = –vcom +vcom = 0
P
O
vcom
T
v = vcom
v = vcom
P
O
T
v = vcom
v = –vcom
+
+
=
=
(c)Rolling motion (b)Pure translation (a)Pure rotation
ω ω
Figure 11-4 Rolling motion of a wheel as a combination of purely rotational motion and
purely translational motion. (a) The purely rotational motion:All points on the wheel
move with the same angular speed v. Points on the outside edge of the wheel all move
with the same linear speed vvcom.The linear velocities of two such points, at top (T)
and bottom (P) of the wheel, are shown. (b) The purely translational motion:All points on
the wheel move to the right with the same linear velocity .(c) The rolling motion of
the wheel is the combination of (a) and (b).
v
:
com
v
:
Figure 11-3 The center of mass Oof a rolling
wheel moves a distance sat velocity
while the wheel rotates through angle u.
The point Pat which the wheel makes
contact with the surface over which the
wheel rolls also moves a distance s.
v
:
com
P P
OO
θ
s
vcom
s
vcom
(at point T) is moving at speed 2vcom, faster than any other portion of the
wheel. These results are demonstrated in Fig. 11-5, which is a time exposure of
a rolling bicycle wheel.You can tell that the wheel is moving faster near its top
than near its bottom because the spokes are more blurred at the top than at
the bottom.
The motion of any round body rolling smoothly over a surface can be sepa-
rated into purely rotational and purely translational motions, as in Figs. 11-4a
and 11-4b.
Rolling as Pure Rotation
Figure 11-6 suggests another way to look at the rolling motion of a wheel
namely, as pure rotation about an axis that always extends through the point
where the wheel contacts the street as the wheel moves. We consider the rolling
motion to be pure rotation about an axis passing through point Pin Fig.11-4cand
perpendicular to the plane of the figure. The vectors in Fig. 11-6 then represent
the instantaneous velocities of points on the rolling wheel.
Question: What angular speed about this new axis will a stationary observer as-
sign to a rolling bicycle wheel?
Answer: The same vthat the rider assigns to the wheel as she or he observes it
in pure rotation about an axis through its center of mass.
To verify this answer, let us use it to calculate the linear speed of the top of the
rolling wheel from the point of view of a stationary observer. If we call the
wheel’s radius R, the top is a distance 2Rfrom the axis through Pin Fig. 11-6, so
the linear speed at the top should be (using Eq. 11-2)
vtop (v)(2R)2(vR)2vcom,
in exact agreement with Fig. 11-4c. You can similarly verify the linear speeds
shown for the portions of the wheel at points Oand Pin Fig.11-4c.
297
11-1 ROLLING AS TRANSLATION AND ROTATION COMBINED
Figure 11-5 A photograph of a rolling
bicycle wheel.The spokes near the
wheel’s top are more blurred than
those near the bottom because the top
ones are moving faster,as Fig. 11-4c
shows.
Courtesy Alice Halliday
Checkpoint 1
The rear wheel on a clown’s bicycle has twice the radius of the front wheel. (a) When
the bicycle is moving, is the linear speed at the very top of the rear wheel greater than,
less than, or the same as that of the very top of the front wheel? (b) Is the angular speed
of the rear wheel greater than, less than, or the same as that of the front wheel?
Figure 11-6 Rolling can be viewed as pure
rotation, with angular speed v, about an
axis that always extends through P.The
vectors show the instantaneous linear
velocities of selected points on the rolling
wheel.You can obtain the vectors by
combining the translational and rotational
motions as in Fig.11-4.
T
Rotation axis at P
O
The Kinetic Energy of Rolling
Let us now calculate the kinetic energy of the rolling wheel as measured by the
stationary observer. If we view the rolling as pure rotation about an axis through
Pin Fig.11-6, then from Eq. 10-34 we have
(11-3)
in which vis the angular speed of the wheel and IPis the rotational inertia of the
wheel about the axis through P. From the parallel-axis theorem of Eq. 10-36
(IIcom Mh2), we have
IPIcom MR2, (11-4)
in which Mis the mass of the wheel, Icom is its rotational inertia about an axis
through its center of mass, and R(the wheel’s radius) is the perpendicular
distance h. Substituting Eq. 11-4 into Eq. 11-3, we obtain
and using the relation vcom vR(Eq. 11-2) yields
(11-5)
We can interpret the term as the kinetic energy associated with the
rotation of the wheel about an axis through its center of mass (Fig.11-4a), and the
term as the kinetic energy associated with the translational motion of the
wheel’s center of mass (Fig. 11-4b).Thus, we have the following rule:
1
2Mvcom
2
1
2Icomv2
K1
2Icomv21
2
v2
com.
K1
2Icomv21
2
R2v2,
K1
2IPv2,
298 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
A smoothly rolling wheel has kinetic energy
where Icom is the rotational inertia of the wheel about its cen-
ter of mass and Mis the mass of the wheel.
If the wheel is being accelerated but is still rolling smoothly,
the acceleration of the center of mass is related to thea
:
com
K1
2Icomv21
2
v2
com,
angular acceleration aabout the center with
acom aR.
If the wheel rolls smoothly down a ramp of angle u, its
acceleration along an xaxis extending up the ramp is
acom, x g sin u
1Icom /MR2.
Key Ideas
A rolling object has two types of kinetic energy: a rotational kinetic energy
due to its rotation about its center of mass and a translational kinetic
(1
2Icomv2)
energy due to translation of its center of mass.(1
2Mvcom
2)
11-2 FORCES AND KINETIC ENERGY OF ROLLING
After reading this module, you should be able to . . .
11.03 Calculate the kinetic energy of a body in smooth rolling as
the sum of the translational kinetic energy of the center of mass
and the rotational kinetic energy around the center of mass.
11.04 Apply the relationship between the work done on a
smoothly rolling object and the change in its kinetic energy.
11.05 For smooth rolling (and thus no sliding), conserve me-
chanical energy to relate initial energy values to the values
at a later point.
11.06 Draw a free-body diagram of an accelerating body that is
smoothly rolling on a horizontal surface or up or down a ramp.
11.07 Apply the relationship between the center-of-mass
acceleration and the angular acceleration.
11.08 For smooth rolling of an object up or down a
ramp, apply the relationship between the objects
acceleration, its rotational inertia, and the angle of
the ramp.
Learning Objectives
The Forces of Rolling
Friction and Rolling
If a wheel rolls at constant speed, as in Fig. 11-3, it has no tendency to slide at the
point of contact P, and thus no frictional force acts there. However, if a net force
acts on the rolling wheel to speed it up or to slow it, then that net force causes ac-
celeration of the center of mass along the direction of travel. It also causes
the wheel to rotate faster or slower, which means it causes an angular
acceleration a.These accelerations tend to make the wheel slide at P.Thus, a fric-
tional force must act on the wheel at Pto oppose that tendency.
If the wheel does not slide, the force is a static frictional force and the
motion is smooth rolling.We can then relate the magnitudes of the linear acceler-
ation and the angular acceleration aby differentiating Eq. 11-2 with respect
to time (with Rheld constant). On the left side, dvcom/dt is acom, and on the right
side dv/dt is a. So, for smooth rolling we have
acom aR(smooth rolling motion). (11-6)
If the wheel does slide when the net force acts on it, the frictional force that
acts at Pin Fig. 11-3 is a kinetic frictional force The motion then is not smooth
rolling, and Eq. 11-6 does not apply to the motion. In this chapter we discuss only
smooth rolling motion.
Figure 11-7 shows an example in which a wheel is being made to rotate faster
while rolling to the right along a flat surface, as on a bicycle at the start of a race.
The faster rotation tends to make the bottom of the wheel slide to the left at
point P. A frictional force at P, directed to the right, opposes this tendency to
slide. If the wheel does not slide, that frictional force is a static frictional force
(as shown), the motion is smooth rolling, and Eq. 11-6 applies to the motion.
(Without friction, bicycle races would be stationary and very boring.)
If the wheel in Fig. 11-7 were made to rotate slower, as on a slowing bicy-
cle, we would change the figure in two ways: The directions of the center-of-
mass acceleration and the frictional force at point Pwould now be to
the left.
Rolling Down a Ramp
Figure 11-8 shows a round uniform body of mass Mand radius Rrolling smoothly
down a ramp at angle u,along an xaxis.We want to find an expression for the body’s
f
:
s
a
:
com
f
:
s
f
:
k.
a
:
com
f
:
s
a
:
com
299
11-2 FORCES AND KINETIC ENERGY OF ROLLING
Figure 11-7 A wheel rolls horizontally with-
out sliding while accelerating with linear
acceleration , as on a bicycle at the start
of a race.A static frictional force acts
on the wheel at P, opposing its tendency
to slide.
f
:
s
a
:
com
Pfs
acom
Figure 11-8 A round uniform body of radius Rrolls down a ramp.The forces that act on it
are the gravitational force F
:
g, a normal force F
:
N, and a frictional force f
:
spointing up the
ramp. (For clarity, vector F
:
Nhas been shifted in the direction it points until its tail is at the
center of the body.)
R
Fgcos
θ
F
g
Fgsin
θ
θ
θ
P
x
fs
FNForces FN and Fg cos
merely balance.
θ
Forces Fg sin and fs
determine the linear
acceleration down
the ramp.
θ
The torque due to fs
determines the
angular acceleration
around the com.
acceleration acom,xdown the ramp.We do this by using Newton’s second law in both
its linear version (Fnet Ma) and its angular version (tnet Ia).
We start by drawing the forces on the body as shown in Fig.11-8:
1. The gravitational force on the body is directed downward. The tail of the
vector is placed at the center of mass of the body. The component along the
ramp is Fgsin u,which is equal to Mg sin u.
2. A normal force is perpendicular to the ramp. It acts at the point of
contact P, but in Fig. 11-8 the vector has been shifted along its direction until
its tail is at the body’s center of mass.
3. A static frictional force acts at the point of contact Pand is directed up
the ramp.(Do you see why? If the body were to slide at P, it would slide down the
ramp.Thus, the frictional force opposing the sliding must be up the ramp.)
We can write Newton’s second law for components along the xaxis in Fig. 11-8
(Fnet,xmax) as
fsMg sin uMacom,x. (11-7)
This equation contains two unknowns, fsand acom,x. (We should not assume that fs
is at its maximum value fs,max. All we know is that the value of fsis just right for
the body to roll smoothly down the ramp, without sliding.)
We now wish to apply Newton’s second law in angular form to the body’s ro-
tation about its center of mass. First, we shall use Eq. 10-41 to write the
torques on the body about that point. The frictional force has moment arm R
and thus produces a torque Rfs, which is positive because it tends to rotate the
body counterclockwise in Fig. 11-8. Forces and have zero moment arms
about the center of mass and thus produce zero torques. So we can write the an-
gular form of Newton’s second law (tnet Ia) about an axis through the body’s
center of mass as
RfsIcoma. (11-8)
This equation contains two unknowns, fsand a.
Because the body is rolling smoothly, we can use Eq. 11-6 (acom aR) to relate
the unknowns acom,xand a. But we must be cautious because here acom,xis negative
(in the negative direction of the xaxis) and ais positive (counterclockwise). Thus
we substitute acom,x/Rfor ain Eq.11-8.Then,solving for fs, we obtain
(11-9)
Substituting the right side of Eq. 11-9 for fsin Eq. 11-7, we then find
(11-10)
We can use this equation to find the linear acceleration acom,xof any body rolling
along an incline of angle uwith the horizontal.
Note that the pull by the gravitational force causes the body to come down
the ramp, but it is the frictional force that causes the body to rotate and thus roll.
If you eliminate the friction (by, say, making the ramp slick with ice or grease) or
arrange for Mg sin uto exceed fs,max, then you eliminate the smooth rolling and
the body slides down the ramp.
acom,x g sin u
1Icom/MR2.
fsIcom
acom,x
R2.
F
:
N
F
:
g
f
:
s
(trF)
f
:
s
F
:
N
F
:
g
300 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Checkpoint 2
Disks Aand Bare identical and roll across a floor with equal speeds.Then disk Arolls
up an incline, reaching a maximum height h, and disk Bmoves up an incline that is
identical except that it is frictionless.Is the maximum height reached by disk Bgreater
than, less than, or equal to h?
301
11-3 THE YO-YO
Doing so, substituting for Icom (from Table 10-2f), and
then solving for vcom give us
4.10 m/s. (Answer)
Note that the answer does not depend on Mor R.
(b) What are the magnitude and direction of the frictional
force on the ball as it rolls down the ramp?
KEY IDEA
Because the ball rolls smoothly, Eq. 11-9 gives the frictional
force on the ball.
Calculations: Before we can use Eq. 11-9, we need the
ball’s acceleration acom,xfrom Eq. 11-10:
Note that we needed neither mass Mnor radius Rto find
acom,x. Thus, any size ball with any uniform mass would have
this smoothly rolling acceleration down a 30.0ramp.
We can now solve Eq. 11-9 as
(Answer)
Note that we needed mass Mbut not radius R. Thus, the
frictional force on any 6.00 kg ball rolling smoothly down
a 30.0ramp would be 8.40 N regardless of the ball’s ra-
dius but would be larger for a larger mass.

2
5(6.00 kg)(3.50 m/s2)8.40 N.
fsIcom
acom,x
R2
2
5MR2acom,x
R2
2
5Macom,x
 (9.8 m/s2) sin 30.0
12
5
3.50 m/s2.
acom,x g sin u
1Icom/MR2 g sin u
12
5MR2/MR2
vcom 2(10
7)gh 2(10
7)(9.8 m/s2)(1.20 m)
2
5MR2
Sample Problem 11.01 Ball rolling down a ramp
A uniform ball, of mass M6.00 kg and radius R, rolls
smoothly from rest down a ramp at angle u30.0(Fig. 11-8).
(a) The ball descends a vertical height h1.20 m to reach the
bottom of the ramp.What is its speed at the bottom?
KEY IDEAS
The mechanical energy Eof the ballEarth system is con-
served as the ball rolls down the ramp.The reason is that the
only force doing work on the ball is the gravitational force, a
conservative force. The normal force on the ball from the
ramp does zero work because it is perpendicular to the
ball’s path. The frictional force on the ball from the ramp
does not transfer any energy to thermal energy because the
ball does not slide (it rolls smoothly).
Thus, we conserve mechanical energy (EfEi):
KfUfKiUi, (11-11)
where subscripts fand irefer to the final values (at the
bottom) and initial values (at rest), respectively.The gravita-
tional potential energy is initially UiMgh (where Mis the
ball’s mass) and finally Uf0.The kinetic energy is initially
Ki0. For the final kinetic energy Kf, we need an addi-
tional idea: Because the ball rolls, the kinetic energy in-
volves both translation and rotation, so we include them
both by using the right side of Eq. 11-5.
Calculations: Substituting into Eq. 11-11 gives us
(11-12)
where Icom is the ball’s rotational inertia about an axis
through its center of mass, vcom is the requested speed at the
bottom, and vis the angular speed there.
Because the ball rolls smoothly, we can use Eq. 11-2 to
substitute vcom/Rfor vto reduce the unknowns in Eq. 11-12.
(1
2Icomv21
2
vcom
2)00Mgh,
Additional examples, video, and practice available at WileyPLUS
11-3 THE YO-YO
After reading this module, you should be able to . . .
11.09 Draw a free-body diagram of a yo-yo moving up or
down its string.
11.10 Identify that a yo-yo is effectively an object that rolls
smoothly up or down a ramp with an incline angle of 90.
11.11 For a yo-yo moving up or down its string, apply the rela-
tionship between the yo-yos acceleration and its rotational
inertia.
11.12 Determine the tension in a yo-yos string as the yo-yo
moves up or down its string.
A yo-yo, which travels vertically up or down a string, can be treated as a wheel rolling along an inclined plane at angle u90.
Learning Objectives
Key Idea
The Yo-Yo
A yo-yo is a physics lab that you can fit in your pocket. If a yo-yo rolls down its
string for a distance h, it loses potential energy in amount mgh but gains kinetic
energy in both translational and rotational forms. As it climbs
back up, it loses kinetic energy and regains potential energy.
In a modern yo-yo, the string is not tied to the axle but is looped around it.
When the yo-yo “hits” the bottom of its string, an upward force on the axle from
the string stops the descent. The yo-yo then spins, axle inside loop, with only
rotational kinetic energy. The yo-yo keeps spinning (“sleeping”) until you “wake
it” by jerking on the string, causing the string to catch on the axle and the yo-yo to
climb back up. The rotational kinetic energy of the yo-yo at the bottom of its
string (and thus the sleeping time) can be considerably increased by throwing the
yo-yo downward so that it starts down the string with initial speeds vcom and vin-
stead of rolling down from rest.
To find an expression for the linear acceleration acom of a yo-yo rolling down
a string, we could use Newton’s second law (in linear and angular forms) just as
we did for the body rolling down a ramp in Fig. 11-8.The analysis is the same ex-
cept for the following:
1. Instead of rolling down a ramp at angle uwith the horizontal, the yo-yo rolls
down a string at angle u90with the horizontal.
2. Instead of rolling on its outer surface at radius R, the yo-yo rolls on an axle of
radius R0(Fig. 11-9a).
3. Instead of being slowed by frictional force , the yo-yo is slowed by the force
on it from the string (Fig.11-9b).
The analysis would again lead us to Eq. 11-10. Therefore, let us just change the
notation in Eq. 11-10 and set u90to write the linear acceleration as
(11-13)
where Icom is the yo-yo’s rotational inertia about its center and Mis its mass. A
yo-yo has the same downward acceleration when it is climbing back up.
acom  g
1Icom/MR2
0
,
T
:f
:
s
(1
2Icomv2)(1
2Mvcom
2)
302 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Figure 11-9 (a) A yo-yo, shown in cross
section. The string, of assumed negligible
thickness, is wound around an axle of
radius R0.(b) A free-body diagram for the
falling yo-yo. Only the axle is shown.
Fg
(a) (b)
R
R0
R0
T
11-4 TORQUE REVISITED
After reading this module, you should be able to . . .
11.13 Identify that torque is a vector quantity.
11.14 Identify that the point about which a torque is
calculated must always be specified.
11.15 Calculate the torque due to a force on a particle by
taking the cross product of the particles position vector
and the force vector, in either unit-vector notation or
magnitude-angle notation.
11.16 Use the right-hand rule for cross products to find the
direction of a torque vector.
In three dimensions, torque is a vector quantity defined
relative to a fixed point (usually an origin); it is
where is a force applied to a particle and is a
position vector locating the particle relative to the fixed
point.
r
:
F
:
t
:r
:F
:,
t
:The magnitude of is given by
where fis the angle between and , is the component
of perpendicular to , and is the moment arm of .
The direction of is given by the right-hand rule for cross
products.
t
:
F
:
r
r
:
F
:F
r
:
F
:
trF sin frF
rF,
t
:
Learning Objectives
Key Ideas
Torque Revisited
In Chapter 10 we defined torque tfor a rigid body that can rotate around a fixed
axis. We now expand the definition of torque to apply it to an individual particle
that moves along any path relative to a fixed point (rather than a fixed axis). The
path need no longer be a circle, and we must write the torque as a vector that may
have any direction. We can calculate the magnitude of the torque with a formula
and determine its direction with the right-hand rule for cross products.
Figure 11-10ashows such a particle at point Ain an xy plane.A single force
in that plane acts on the particle, and the particle’s position relative to the origin
Ois given by position vector .The torque acting on the particle relative to the
fixed point Ois a vector quantity defined as
(torque defined). (11-14)
We can evaluate the vector (or cross) product in this definition of by using
the rules in Module 3-3. To find the direction of , we slide the vector (without
changing its direction) until its tail is at the origin O, so that the two vectors in the
vector product are tail to tail as in Fig. 11-10b. We then use the right-hand rule in
Fig. 3-19a, sweeping the fingers of the right hand from (the first vector in the
product) into (the second vector). The outstretched right thumb then gives the
direction of . In Fig.11-10b, it is in the positive direction of the zaxis.t
:
F
:r
:
F
:
t
:
t
:
t
:r
:F
:
t
:
r
:
F
:
t
:
303
11-4 TORQUE REVISITED
Checkpoint 3
The position vector of a particle points along the positive direction of a zaxis. If
the torque on the particle is (a) zero,(b) in the negative direction of x,and (c) in the
negative direction of y,in what direction is the force causing the torque?
r
:
To determine the magnitude of , we apply the general result of Eq. 3-27
(cab sin f), finding trF sin f, (11-15)
where fis the smaller angle between the directions of and when the vectors
are tail to tail. From Fig.11-10b, we see that Eq. 11-15 can be rewritten as
(11-16)
where is the component of perpendicular to r
:. From Fig. 11-10c,
we see that Eq. 11-15 can also be rewritten as
(11-17)
where is the moment arm of (the perpendicular distance
between Oand the line of action of F
:).
F
:
r(r sin f)
trF,
F
:
F
(F sin f)
trF
,
F
:
r
:
t
:
Figure 11-10 Defining torque.(a) A force , lying in an xy plane, acts on a particle at point A.(b) This force produces a torque
on the particle with respect to the origin O.By the right-hand rule for vector (cross) products,the torque vector
points in the positive direction of z.Its magnitude is given by in (b) and by in (c).rFrF
t
:
(r
:F
:)
F
:
A
φ
φ
z
x
y
F
(= r× F)
F (redrawn, with
tail at origin)
(b)
O
φ
τ
F
r
A
z
x
y
r
Line of action of F
φ
(c)
O
τ
F
r
A
φ
z
x
y
(a)
O
F
r
Cross r into F.
Torque is in the
positive z direction.
τ
304 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
the position vector are to see.) In Fig. 11-11d, the angle be-
tween the directions of and is 90and the symbol F
:
3
r
:
Sample Problem 11.02 Torque on a particle due to a force
In Fig. 11-11a, three forces, each of magnitude 2.0 N, act on a
particle. The particle is in the xz plane at point Agiven by
position vector , where r3.0 m and u30. What is the
torque, about the origin O, due to each force?
KEY IDEA
Because the three force vectors do not lie in a plane, we
must use cross products, with magnitudes given by Eq. 11-15
(trF sin f) and directions given by the right-hand rule.
Calculations: Because we want the torques with respect to
the origin O, the vector required for each cross product
is the given position vector. To determine the angle fbe-
tween and each force, we shift the force vectors of Fig.11-
11a, each in turn, so that their tails are at the origin. Figures
11-11b,c, and d, which are direct views of the xz plane, show
the shifted force vectors and , respectively. (Note
how much easier the angles between the force vectors and
F
:
3
F
:
2,F
:
1,
r
:
r
:
r
:
Figure 11-11 (a) A particle at point Ais acted on by three forces,each parallel to a coordinate axis.The angle f(used in finding torque) is shown
(b) for and (c) for .(d) Torque t
:
3is perpendicular to both and (force is directed into the plane of the figure).(e) The torques.F
:
3
F
:
3
r
:
F
:
2
F
:
1
Additional examples, video, and practice available at WileyPLUS
A
z
x
y
(a)
F3F1
F2
θ
O
r
x
z
θ
= 30°
(b)
φ
1= 150°
OF1
r
x
z
F1
r
x
z
F1
r
Cross into F
1
. Torque
1
is into the figure (negative y).
r
τ
x
z
θ
= 30°
φ
2= 120°
O
(c)F2
r
x
z
F2
r
F
2
F
F
F
2
FF
F
F
F
F
2
FF
Cross into
Torque 2 is out of
the figure (positive y).
F2.
r
τ
x
z
F2
r
x
z
θ
= 30°
τ
3
θ
(d)
F3
r
x
z
τ
3
r
3
r
3
r
3
x
z
τ
3
r
3
r
r
3
3
3
r
y
z
x
2
1
O
τ
3
θ
(e)
τ
τ
The three
torques.
Cross into
Torque 3 is
in the xz plane.
F3.
r
τ
A
means is directed into the page. (For out of the page, weF
:
3
would use .)
Now, applying Eq. 11-15, we find
t1rF1sin f1(3.0 m)(2.0 N)(sin 150)3.0 Nm,
t2rF2sin f2(3.0 m)(2.0 N)(sin 120)5.2 Nm,
and t3rF3sin f3(3.0 m)(2.0 N)(sin 90)
6.0 Nm. (Answer)
Next, we use the right-hand rule, placing the fingers of
the right hand so as to rotate into through the smaller of
the two angles between their directions. The thumb points in
the direction of the torque.Thus t
:
1is directed into the page in
Fig. 11-11b;t
:
2is directed out of the page in Fig. 11-11c; and
t
:
3is directed as shown in Fig. 11-11d. All three torque vec-
tors are shown in Fig.11-11e.
F
:
r
:
305
11-5 ANGULAR MOMENTUM
11-5 ANGULAR MOMENTUM
After reading this module, you should be able to . . .
11.17 Identify that angular momentum is a vector quantity.
11.18 Identify that the fixed point about which an angular
momentum is calculated must always be specified.
11.19 Calculate the angular momentum of a particle by taking
the cross product of the particle’s position vector and its
momentum vector, in either unit-vector notation or
magnitude-angle notation.
11.20 Use the right-hand rule for cross products to find the
direction of an angular momentum vector.
The angular momentum of a particle with linear momen-
tum , mass m, and linear velocity is a vector quantity
defined relative to a fixed point (usually an origin) as
The magnitude of is given by
rprmv,
rprmv
rmv sin f
:
:r
:p
:m(r
:v
:
).
v
:
p
:
:where fis the angle between and , and are
the components of and perpendicular to , and is
the perpendicular distance between the fixed point and
the extension of .
The direction of is given by the right-hand rule: Position
your right hand so that the fingers are in the direction of .
Then rotate them around the palm to be in the direction of .
Your outstretched thumb gives the direction of .
:p
:
r
:
:
p
:
r
r
:
v
:
p
:
v
p
p
:
r
:
Learning Objectives
Key Ideas
Angular Momentum
Recall that the concept of linear momentum and the principle of conservation
of linear momentum are extremely powerful tools. They allow us to predict
the outcome of, say, a collision of two cars without knowing the details of the col-
lision. Here we begin a discussion of the angular counterpart of , winding up in
Module 11-8 with the angular counterpart of the conservation principle, which
can lead to beautiful (almost magical) feats in ballet, fancy diving,ice skating, and
many other activities.
Figure 11-12 shows a particle of mass mwith linear momentum as
it passes through point Ain an xy plane. The angular momentum of this parti-
cle with respect to the origin Ois a vector quantity defined as
(angular momentum defined), (11-18)
where is the position vector of the particle with respect to O. As the particle
moves relative to Oin the direction of its momentum , position vector
rotates around O. Note carefully that to have angular momentum about O, the
particle does not itself have to rotate around O. Comparison of Eqs. 11-14 and 11-18
shows that angular momentum bears the same relation to linear momentum that
torque does to force. The SI unit of angular momentum is the kilogram-
meter-squared per second (kg m2/s), equivalent to the joule-second (Js).
Direction. To find the direction of the angular momentum vector in Fig. 11-
12, we slide the vector until its tail is at the origin O.Then we use the right-hand
rule for vector products, sweeping the fingers from into . The outstretched
thumb then shows that the direction of is in the positive direction of the zaxis in
Fig.11-12.This positive direction is consistent with the counterclockwise rotation of
position vector about the zaxis, as the particle moves. (A negative direction of
would be consistent with a clockwise rotation of about the zaxis.)
Magnitude. To find the magnitude of ,we use the general result of Eq. 3-27
to write (11-19)
where fis the smaller angle between and when these two vectors are tailp
:
r
:
rmv sin f,
:r
:
:
r
:
:p
:
r
:
p
:
:
r
:
p
:
(mv
:
)
r
:
:r
:p
:m(r
:v
:
)
:p
:(mv
:)
p
:
p
:
Figure 11-12 Defining angular momentum.A
particle passing through point Ahas linear
momentum , with the vector
lying in an xy plane.The particle has angular
momentum with respect to the
origin O. By the right-hand rule,the angular
momentum vector points in the positive
direction of z.(a) The magnitude of is
given by .(b) The magni-rprmv
:
:(r
:p
:)
p
:
p
:(mv
:)
A
φ
φ
z
x
y
p
(= r
×
p)
p (redrawn, with
tail at origin)
A
z
x
y
Extension of p
φ
(a)
(b)
O
O
φ
r
r
rp
p
tude of is also given by .rprmv
:
to tail. From Fig.11-12a, we see that Eq. 11-19 can be rewritten as
(11-20)
where is the component of perpendicular to and is the component
of perpendicular to . From Fig. 11-12b, we see that Eq. 11-19 can also be
rewritten as
(11-21)
where is the perpendicular distance between Oand the extension of .
Important. Note two features here: (1) angular momentum has meaning
only with respect to a specified origin and (2) its direction is always perpendicu-
lar to the plane formed by the position and linear momentum vectors and .p
:
r
:
p
:
r
rprmv,
r
:
v
:
v
r
:
p
:
p
rprmv,
306 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Checkpoint 4
In part aof the figure, particles 1 and 2 move around point Oin circles
with radii 2 m and 4 m. In part b, particles 3 and 4 travel along straight
lines at perpendicular distances of 4 m and 2 m from point O. Particle 5
moves directly away from O.All five particles have the same mass and
the same constant speed. (a) Rank the particles according to the magni-
tudes of their angular momentum about point O, greatest first. (b)
Which particles have negative angular momentum about point O?
(a)(b)
O
2
13
5
4
O
around Oas particle 1 moves. Thus, the angular momen-
tum vector for particle 1 is
Similarly, the magnitude of is
and the vector product is into the page, which is the
negative direction, consistent with the clockwise rotation of
around Oas particle 2 moves. Thus, the angular momen-
tum vector for particle 2 is
The net angular momentum for the two-particle system is
(Answer)
The plus sign means that the system’s net angular momen-
tum about point Ois out of the page.
2.0 kgm2/s.
L1210 kgm2/s (8.0 kgm2/s)
28.0 kgm2/s.
r
:
2
r2
:p2
:
8.0 kgm2/s,
2r2p2(4.0 m)(2.0 kgm/s)
:
2
110 kgm2/s.
r
:
1
Sample Problem 11.03 Angular momentum of a two-particle system
Figure 11-13 shows an overhead view of two particles moving
at constant momentum along horizontal paths. Particle 1, with
momentum magnitude p15.0 kgm/s, has position vector
and will pass 2.0 m from point O. Particle 2, with momentum
magnitude p22.0 kgm/s, has position vector and will pass
4.0 m from point O. What are the magnitude and direction of
the net angular momentum about point Oof the two-
particle system?
KEY IDEA
To find , we can first find the individual angular momenta
and and then add them.To evaluate their magnitudes,
we can use any one of Eqs. 11-18 through 11-21. However,
Eq. 11-21 is easiest, because we are given the perpendicular
distances and and the momen-
tum magnitudes p1and p2.
Calculations: For particle 1, Eq. 11-21 yields
To find the direction of vector , we use Eq. 11-18 and the
right-hand rule for vector products. For , the vector
product is out of the page, perpendicular to the plane of
Fig. 11-13. This is the positive direction, consistent with the
counterclockwise rotation of the particle’s position vector
r1
:p1
:
:
1
10 kgm2/s.
1r1p1(2.0 m)(5.0 kgm/s)
r2 (4.0 m)r1 (2.0 m)
:
2
:
1
L
:
L
:
r
:
2
r
:
1
Figure 11-13 Two particles pass
near point O.
r1
r2
r2
r1
O
p2
p1
Additional examples, video, and practice available at WileyPLUS
307
11-6 NEWTON’S SECOND LAW IN ANGULAR FORM
11-6 NEWTON’S SECOND LAW IN ANGULAR FORM
After reading this module, you should be able to . . .
11.21 Apply Newton’s second law in angular form to relate the torque acting on a particle to the resulting rate of change of the
particle’s angular momentum, all relative to a specified point.
Newton’s second law for a particle can be written in angular form as
where is the net torque acting on the particle and is the angular momentum of the particle.
:
t
:
net
t
:
net d
:
dt ,
Learning Objective
Key Idea
Newton’s Second Law in Angular Form
Newton’s second law written in the form
(single particle) (11-22)
expresses the close relation between force and linear momentum for a single
particle. We have seen enough of the parallelism between linear and angular
quantities to be pretty sure that there is also a close relation between torque
and angular momentum. Guided by Eq. 11-22, we can even guess that it must be
(single particle). (11-23)
Equation 11-23 is indeed an angular form of Newton’s second law for a single particle:
t
:
net d
:
dt
F
:
net dp
:
dt
The (vector) sum of all the torques acting on a particle is equal to the time rate of
change of the angular momentum of that particle.
Equation 11-23 has no meaning unless the torques and the angular momentum
are defined with respect to the same point, usually the origin of the coordinate
system being used.
Proof of Equation 11-23
We start with Eq. 11-18, the definition of the angular momentum of a particle:
where is the position vector of the particle and is the velocity of the particle.
Differentiating* each side with respect to time tyields
(11-24)
However, is the acceleration of the particle, and is its velocity .
Thus, we can rewrite Eq. 11-24 as
d
:
dt m(r
:a
:v
:v
:
).
v
:
dr
:
/dta
:
dv
:
/dt
d
:
dt m
r
:dv
:
dt dr
:
dt v
:
.
v
:
r
:
:m(r
:v
:),
:t
:
*In differentiating a vector product, be sure not to change the order of the two quantities (here and
) that form that product. (See Eq. 3-25.)
v
:
r
:
308 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Checkpoint 5
The figure shows the position vector of a particle
at a certain instant, and four choices for the direc-
tion of a force that is to accelerate the particle.All
four choices lie in the xy plane. (a) Rank the
choices according to the magnitude of the time rate
of change they produce in the angular mo-
mentum of the particle about point O, greatest
first. (b) Which choice results in a negative rate of change about O?
(d
:/dt)
r
:
x
F
1
O
y
F
2
F
3
F
4
r
on a particle and the angular momentum of the particle
are calculated around the same point, then the torque is re-
lated to angular momentum by Eq. 11-23 ( ).
Calculations: In order to use Eq. 11-18 to find the angular
momentum about the origin, we first must find an expres-
sion for the particle’s velocity by taking a time derivative of
its position vector. Following Eq. 4-10 ( ), we write
,
with in meters per second.
Next, let’s take the cross product of and using the
template for cross products displayed in Eq. 3-27:
Here the generic is and the generic is . However,
because we really don’t want to do more work than
needed, let’s first just think about our substitutions into
v
:
b
:
r
:
a
:
a
:b
:(aybzbyaz)i
ˆ(azbxbzax)j
ˆ(axbybxay)k
ˆ .
v
:
r
:
v
:
(4.00t1.00)i
ˆ
v
:d
dt ((2.00t2t)i
ˆ5.00j
ˆ)
dr
:
/dtv
:
d
:
/dtt
:
Sample Problem 11.04 Torque and the time derivative of angular momentum
Figure 11-14ashows a freeze-frame of a 0.500 kg particle
moving along a straight line with a position vector given by
,
with in meters and tin seconds, starting at t0. The posi-
tion vector points from the origin to the particle. In unit-vector
notation, find expressions for the angular momentum of
the particle and the torque acting on the particle, both
with respect to (or about) the origin. Justify their algebraic
signs in terms of the particle’s motion.
KEY IDEAS
(1) The point about which an angular momentum of a par-
ticle is to be calculated must always be specified. Here it is
the origin. (2) The angular momentum of a particle is
given by Eq. 11-18 . (3) The sign
associated with a particle’s angular momentum is set by the
sense of rotation of the particle’s position vector (around
the rotation axis) as the particle moves: clockwise is nega-
tive and counterclockwise is positive. (4) If the torque acting
(
:r
:p
:m(r
:v
:))
:
t
:
:
r
:
r
:(2.00t2t)i
ˆ5.00j
ˆ
Now (the vector product of any vector with itself is zero because the
angle between the two vectors is necessarily zero). Thus, the last term of this
expression is eliminated and we then have
We now use Newton’s second law to replace with its equal, the
vector sum of the forces that act on the particle, obtaining
(11-25)
Here the symbol indicates that we must sum the vector products for all
the forces. However, from Eq. 11-14, we know that each one of those vector prod-
ucts is the torque associated with one of the forces.Therefore, Eq. 11-25 tells us that
This is Eq. 11-23, the relation that we set out to prove.
t
:
net d
:
dt .
r
:F
:
d
:
dt r
:F
:
net
(r
:F
:).
ma
:
(F
:
net ma
:)
d
:
dt m(r
:a
:)r
:ma
:.
v
:v
:0
309
11-6 NEWTON’S SECOND LAW IN ANGULAR FORM
Figure 11-14 (a) A particle moving in a straight line, shown at time
t0. (b) The position vector at t0,1.00 s, and 2.00 s. (c) The first
step in applying the right-hand rule for cross products.(d) The sec-
ond step. (e) The angular momentum vector and the torque vector
are along the zaxis,which extends out of the plane of the figure.
y(m)
5
(a)
v
x(m)
y(m)
t2 s t1 s t0
10 3
(b)
x(m)
r2
r1r0
y
(c)
r
x
y
(d)
v
x
y
(e)
x
t
Both angular momentum
and torque point out of
figure, in the positive z
direction.
Additional examples, video, and practice available at WileyPLUS
the generic cross product. Because lacks any zcomponent
and because lacks any yor zcomponent, the only nonzero
term in the generic cross product is the very last one .
So,let’s cut to the (mathematical) chase by writing
Note that, as always, the cross product produces a vector
that is perpendicular to the original vectors.
To finish up Eq. 11-18, we multiply by the mass, finding
(Answer)
The torque about the origin then immediately follows from
Eq. 11-23:
(Answer)
which is in the positive direction of the zaxis.
Our result for tells us that the angular momentum is
in the positive direction of the zaxis. To make sense of that
positive result in terms of the rotation of the position vector,
:
10.0k
ˆ kgm2/s210.0k
ˆ Nm,
t
:d
dt (10.0t2.50)k
ˆ kgm2/s
(10.0t2.50)k
ˆ kgm2/s.
:(0.500 kg)[(20.0t5.00)k
ˆ m2/s]
r
:v
:(4.00t1.00)(5.00)k
ˆ(20.0t5.00)k
ˆ m2/s.
(bxay)k
ˆ
v
:
r
:
let’s evaluate that vector for several times:
t0,
t1.00 s,
t2.00 s,
By drawing these results as in Fig. 11-14b, we see that ro-
tates counterclockwise in order to keep up with the particle.
That is the positive direction of rotation. Thus, even though
the particle is moving in a straight line, it is still moving
counterclockwise around the origin and thus has a positive
angular momentum.
We can also make sense of the direction of by applying
the right-hand rule for cross products (here , or, if you
like, , which gives the same direction). For any mo-
ment during the particle’s motion, the fingers of the right
hand are first extended in the direction of the first vector in
the cross product ( ) as indicated in Fig. 11-14c. The orienta-
tion of the hand (on the page or viewing screen) is then ad-
justed so that the fingers can be comfortably rotated about
the palm to be in the direction of the second vector in the
cross product ( ) as indicated in Fig. 11-14d.The outstretched
thumb then points in the direction of the result of the cross
product. As indicated in Fig. 11-14e, the vector is in the posi-
tive direction of the zaxis (which is directly out of the plane
of the figure), consistent with our previous result. Figure
11-14ealso indicates the direction of , which is also in the
positive direction of the zaxis because the angular momen-
tum is in that direction and is increasing in magnitude.
t
:
v
:
r
:
mr
:v
:
r
:v
:
:
r
:
10.0i
ˆ5.00j
ˆ mr
:
2
3.00i
ˆ5.00j
ˆ mr
:
1
5.00j
ˆ mr
:
0
The Angular Momentum of a System of Particles
Now we turn our attention to the angular momentum of a system of particles with
respect to an origin. The total angular momentum of the system is the (vector)
sum of the angular momenta of the individual particles (here with label i):
(11-26)
With time, the angular momenta of individual particles may change because
of interactions between the particles or with the outside.We can find the resulting
change in by taking the time derivative of Eq. 11-26.Thus,
(11-27)
From Eq. 11-23, we see that is equal to the net torque on the ith
particle.We can rewrite Eq. 11-27 as
(11-28)
That is, the rate of change of the system’s angular momentum is equal to the
vector sum of the torques on its individual particles. Those torques include inter-
nal torques (due to forces between the particles) and external torques (due to
forces on the particles from bodies external to the system). However, the forces
between the particles always come in third-law force pairs so their torques sum to
zero. Thus, the only torques that can change the total angular momentum of
the system are the external torques acting on the system.
Net External Torque. Let represent the net external torque, the vector
sum of all external torques on all particles in the system. Then we can write
Eq. 11-28 as
(system of particles), (11-29)
:
net dL
:
dt
:
net
L
:
L
:
dL
:
dt
n
i1
:
net,i.
:
net,i
d
:
i/dt
dL
:
dt
n
i1
di
:
dt .
L
:
L
:
:
1
:
2
:
3 
:
n
n
i1
:
i.
:L
:
310 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
11-7 ANGULAR MOMENTUM OF A RIGID BODY
After reading this module, you should be able to . . .
11.22 For a system of particles, apply Newton’s second law
in angular form to relate the net torque acting on the
system to the rate of the resulting change in the system’s
angular momentum.
11.23 Apply the relationship between the angular momentum
of a rigid body rotating around a fixed axis and the body’s
rotational inertia and angular speed around that axis.
11.24 If two rigid bodies rotate about the same axis, calculate
their total angular momentum.
The angular momentum of a system of particles is
the vector sum of the angular momenta of the individual
particles:
The time rate of change of this angular momentum is equal
to the net external torque on the system (the vector sum of
L
:
:
1
:
2
:
n
n
i1
:
i.
L
:the torques due to interactions of the particles of the system
with particles external to the system):
(system of particles).
For a rigid body rotating about a fixed axis, the component
of its angular momentum parallel to the rotation axis is
LIv(rigid body, fixed axis).
t
:
net dL
:
dt
Learning Objectives
Key Ideas
311
11-7 ANGULAR MOMENTUM OF A RIGID BODY
The net external torque acting on a system of particles is equal to the time
rate of change of the system’s total angular momentum .L
:
:
net
Equation 11-29 is analogous to (Eq. 9-27) but requires extra
caution:Torques and the system’s angular momentum must be measured relative
to the same origin. If the center of mass of the system is not accelerating relative
to an inertial frame, that origin can be any point. However, if it is accelerating,
then it must be the origin. For example, consider a wheel as the system of parti-
cles. If it is rotating about an axis that is fixed relative to the ground, then the ori-
gin for applying Eq. 11-29 can be any point that is stationary relative to the
ground. However, if it is rotating about an axis that is accelerating (such as when
it rolls down a ramp), then the origin can be only at its center of mass.
The Angular Momentum of a Rigid Body
Rotating About a Fixed Axis
We next evaluate the angular momentum of a system of particles that form a rigid
body that rotates about a fixed axis. Figure 11-15ashows such a body.The fixed axis
of rotation is a zaxis, and the body rotates about it with constant angular speed v.
We wish to find the angular momentum of the body about that axis.
We can find the angular momentum by summing the zcomponents of the an-
gular momenta of the mass elements in the body. In Fig. 11-15a, a typical mass el-
ement, of mass mi, moves around the zaxis in a circular path.The position of the
mass element is located relative to the origin Oby position vector . The radius
of the mass element’s circular path is the perpendicular distance between the
element and the zaxis.
The magnitude of the angular momentum of this mass element, with
respect to O, is given by Eq. 11-19:
where piand viare the linear momentum and linear speed of the mass element,
and 90is the angle between and . The angular momentum vector for the
mass element in Fig. 11-15ais shown in Fig. 11-15b; its direction must be perpen-
dicular to those of and .
The z Components. We are interested in the component of that is parallel
to the rotation axis, here the zaxis.That zcomponent is
The zcomponent of the angular momentum for the rotating rigid body as a
whole is found by adding up the contributions of all the mass elements that make
up the body.Thus, because we may write
(11-30)
We can remove vfrom the summation here because it has the same value for all
points of the rotating rigid body.
The quantity in Eq. 11-30 is the rotational inertia Iof the body
about the fixed axis (see Eq. 10-33).Thus Eq.11-30 reduces to
LIv(rigid body, fixed axis). (11-31)
miri
2
v
n
i1
miri
2
.
Lz
n
i1
iz
n
i1
mivir
i
n
i1
mi(vr
i)r
i
vvr,
iz i sin u(ri sin u)(mivi)rimivi.
:
i
p
:
i
r
:
i
:
i
p
:
i
r
:
i
i(ri)( pi)(sin 90)(ri)(mivi),
:
i
ri,
r
:
i
F
:
net dP
:
/dt
Figure 11-15 (a) A rigid body rotates about a
zaxis with angular speed v.A mass ele-
ment of mass miwithin the body moves
about the zaxis in a circle with radius .
The mass element has linear momentum
and it is located relative to the origin
Oby position vector .Here the mass
element is shown when is parallel to the
xaxis. (b) The angular momentum with
respect to O,of the mass element in (a).The
zcomponent is also shown.iz
:
i,
ri
r
:
i
p
:
i,
ri
θ
z
x
y
Δ
mi
ri
ri
pi
O
z
x
y
i
θ
θ
iz
O
(a)
(b)
which is Newton’s second law in angular form.It says:
Conservation of Angular Momentum
So far we have discussed two powerful conservation laws, the conservation of
energy and the conservation of linear momentum. Now we meet a third law of
this type, involving the conservation of angular momentum. We start from
We have dropped the subscript z, but you must remember that the angular
momentum defined by Eq. 11-31 is the angular momentum about the rotation
axis.Also, Iin that equation is the rotational inertia about that same axis.
Table 11-1, which supplements Table 10-3, extends our list of corresponding
linear and angular relations.
312 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Checkpoint 6
In the figure, a disk, a
hoop, and a solid sphere
are made to spin about
fixed central axes (like a
top) by means of strings
wrapped around them, with the strings producing the same constant tangential force
on all three objects.The three objects have the same mass and radius,and they are
initially stationary. Rank the objects according to (a) their angular momentum about
their central axes and (b) their angular speed, greatest first, when the strings have
been pulled for a certain time t.
F
:
Disk Hoop Sphere
F F F
11-8 CONSERVATION OF ANGULAR MOMENTUM
After reading this module, you should be able to . . .
11.25 When no external net torque acts on a system along a specified axis, apply the conservation of angular momentum to
relate the initial angular momentum value along that axis to the value at a later instant.
The angular momentum of a system remains constant if the net external torque acting on the system is zero:
(isolated system)
or (isolated system).
This is the law of conservation of angular momentum.
L
:
iL
:
f
L
:a constant
L
:
Learning Objective
Key Idea
Table 11-1 More Corresponding Variables and Relations for Translational
and Rotational Motiona
Translational Rotational
Force Torque
Linear momentum Angular momentum
Linear momentumbAngular momentumbL
:(
:
i)P
:(p
:
i)
: (r
:p
:)p
:
: (r
:F
:)F
:
Linear momentumbAngular momentumcLIv
Newton’s second lawbNewton’s second lawb
Conservation lawdConservation lawd
aSee also Table 10-3.
bFor systems of particles,including rigid bodies.
cFor a rigid body about a fixed axis,with Lbeing the component along that axis.
dFor a closed, isolated system.
L
:a constantP
:a constant
:
net dL
:
dt
F
:
net dP
:
dt
P
:Mv
:
com
313
11-8 CONSERVATION OF ANGULAR MOMENTUM
If the net external torque acting on a system is zero, the angular momentum of
the system remains constant, no matter what changes take place within the system.
L
:
Equations 11-32 and 11-33 are vector equations; as such, they are equivalent
to three component equations corresponding to the conservation of angular
momentum in three mutually perpendicular directions. Depending on the
torques acting on a system, the angular momentum of the system might be con-
served in only one or two directions but not in all directions:
Eq. 11-29 ,which is Newton’s second law in angular form. If no (t
:
net dL
:/dt)
net external torque acts on the system, this equation becomes or
(isolated system). (11-32)
This result, called the law of conservation of angular momentum, can also be
written as
,
or (isolated system). (11-33)
Equations 11-32 and 11-33 tell us:
L
:
iL
:
f
net angular momentum
at some initial time ti
net angular momentum
at some later time tf
L
:a constant
dL
:/dt 0,
If the component of the net external torque on a system along a certain axis is
zero, then the component of the angular momentum of the system along that axis
cannot change, no matter what changes take place within the system.
This is a powerful statement: In this situation we are concerned with only the initial
and final states of the system; we do not need to consider any intermediate state.
We can apply this law to the isolated body in Fig. 11-15, which rotates around
the zaxis. Suppose that the initially rigid body somehow redistributes its mass
relative to that rotation axis, changing its rotational inertia about that axis.
Equations 11-32 and 11-33 state that the angular momentum of the body cannot
change. Substituting Eq. 11-31 (for the angular momentum along the rotational
axis) into Eq. 11-33, we write this conservation law as
IiviIfvf. (11-34)
Here the subscripts refer to the values of the rotational inertia Iand angular
speed vbefore and after the redistribution of mass.
Like the other two conservation laws that we have discussed, Eqs. 11-32 and
11-33 hold beyond the limitations of Newtonian mechanics. They hold for parti-
cles whose speeds approach that of light (where the theory of special relativity
reigns), and they remain true in the world of subatomic particles (where quantum
physics reigns). No exceptions to the law of conservation of angular momentum
have ever been found.
We now discuss four examples involving this law.
1. The spinning volunteer Figure 11-16 shows a student seated on a stool that
can rotate freely about a vertical axis. The student, who has been set into
rotation at a modest initial angular speed vi, holds two dumbbells in his
outstretched hands. His angular momentum vector lies along the vertical ro-
tation axis, pointing upward.
The instructor now asks the student to pull in his arms; this action reduces
his rotational inertia from its initial value Iito a smaller value Ifbecause he
moves mass closer to the rotation axis. His rate of rotation increases markedly,
L
:
Figure 11-16 (a) The student has a relatively
large rotational inertia about the rotation
axis and a relatively small angular speed.
(b) By decreasing his rotational inertia, the
student automatically increases his angular
speed.The angular momentum of the
rotating system remains unchanged.
L
:
L
Ii
If
Rotation axis
(a)
(b)
i
ω
f
ω
L
314 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
from vito vf.The student can then slow down by extending his arms once more,
moving the dumbbells outward.
No net external torque acts on the system consisting of the student, stool,
and dumbbells.Thus, the angular momentum of that system about the rotation
axis must remain constant, no matter how the student maneuvers the dumb-
bells. In Fig. 11-16a, the student’s angular speed viis relatively low and his ro-
tational inertia Iiis relatively high.According to Eq. 11-34, his angular speed in
Fig. 11-16bmust be greater to compensate for the decreased If.
2. The springboard diver Figure 11-17 shows a diver doing a forward one-and-a-
half-somersault dive. As you should expect, her center of mass follows a para-
bolic path. She leaves the springboard with a definite angular momentum
about an axis through her center of mass, represented by a vector pointing into
the plane of Fig. 11-17, perpendicular to the page. When she is in the air, no net
external torque acts on her about her center of mass, so her angular momentum
about her center of mass cannot change. By pulling her arms and legs into the
closed tuck position, she can considerably reduce her rotational inertia about the
same axis and thus, according to Eq. 11-34, considerably increase her angular
speed. Pulling out of the tuck position (into the open layout position) at the end
of the dive increases her rotational inertia and thus slows her rotation rate so she
can enter the water with little splash. Even in a more complicated dive involving
both twisting and somersaulting, the angular momentum of the diver must be
conserved,in both magnitude and direction, throughout the dive.
3. Long jump When an athlete takes off from the ground in a running long jump,
the forces on the launching foot give the athlete an angular momentum with a
forward rotation around a horizontal axis. Such rotation would not allow the
jumper to land properly: In the landing,the legs should be together and extended
forward at an angle so that the heels mark the sand at the greatest distance. Once
airborne, the angular momentum cannot change (it is conserved) because no ex-
ternal torque acts to change it. However,the jumper can shift most of the angular
momentum to the arms by rotating them in windmill fashion (Fig. 11-18). Then
the body remains upright and in the proper orientation for landing.
L
:
Figure 11-18 Windmill motion of the arms
during a long jump helps maintain body
orientation for a proper landing.
(a) (b)
θ
Figure 11-19 (a) Initial phase of a tour jeté:
large rotational inertia and small angular
speed. (b) Later phase: smaller rotational
inertia and larger angular speed.
Figure 11-17 The diver’s angular momentum
is constant throughout the dive, being
represented by the tail of an arrow that
is perpendicular to the plane of the figure.
Note also that her center of mass (see the
dots) follows a parabolic path.
L
:
L
L
Her angular momentum
is fixed but she can still
control her spin rate.
4. Tour jeté In a tour jeté, a ballet performer leaps with a small twisting motion
on the floor with one foot while holding the other leg perpendicular to the
body (Fig. 11-19a).The angular speed is so small that it may not be perceptible
315
11-8 CONSERVATION OF ANGULAR MOMENTUM
Checkpoint 7
A rhinoceros beetle rides the rim of a small disk that rotates like a merry-go-round.
If the beetle crawls toward the center of the disk, do the following (each relative to
the central axis) increase, decrease, or remain the same for the beetle–disk system:
(a) rotational inertia, (b) angular momentum, and (c) angular speed?
to the audience. As the performer ascends, the outstretched leg is brought
down and the other leg is brought up, with both ending up at angle uto the
body (Fig. 11-19b). The motion is graceful, but it also serves to increase the
rotation because bringing in the initially outstretched leg decreases the per-
former’s rotational inertia. Since no external torque acts on the airborne
performer, the angular momentum cannot change. Thus, with a decrease in
rotational inertia, the angular speed must increase. When the jump is well
executed, the performer seems to suddenly begin to spin and rotates 180
before the initial leg orientations are reversed in preparation for the landing.
Once a leg is again outstretched, the rotation seems to vanish.
Sample Problem 11.05 Conservation of angular momentum, rotating wheel demo
Figure 11-20ashows a student, again sitting on a stool that
can rotate freely about a vertical axis. The student, initially
at rest, is holding a bicycle wheel whose rim is loaded with
lead and whose rotational inertia Iwh about its central axis is
1.2 kgm2. (The rim contains lead in order to make the value
of Iwh substantial.)
The wheel is rotating at an angular speed vwh of 3.9
rev/s; as seen from overhead, the rotation is counterclock-
wise. The axis of the wheel is vertical, and the angular
momentum of the wheel points vertically upward.
The student now inverts the wheel (Fig. 11-20b) so that,
as seen from overhead, it is rotating clockwise. Its angular
momentum is now . The inversion results in the stu-
dent, the stool, and the wheel’s center rotating together as a
composite rigid body about the stool’s rotation axis, with ro-
tational inertia Ib6.8 kgm2. (The fact that the wheel is
also rotating about its center does not affect the mass distri-
bution of this composite body; thus, Ibhas the same value
whether or not the wheel rotates.) With what angular speed
vband in what direction does the composite body rotate af-
ter the inversion of the wheel?
KEY IDEAS
1. The angular speed vbwe seek is related to the final angu-
lar momentum of the composite body about the
stool’s rotation axis by Eq.11-31 (LIv).
2. The initial angular speed vwh of the wheel is related to the
angular momentum of the wheel’s rotation about its
center by the same equation.
3. The vector addition of and gives the total angular
momentum of the system of the student, stool, and
wheel.
4. As the wheel is inverted, no net external torque acts on
L
:
tot
L
:
wh
L
:
b
L
:
wh
L
:
b
L
:
wh
L
:
wh
Figure 11-20 (a) A student holds a bicycle wheel rotating around a
vertical axis.(b) The student inverts the wheel, setting himself
into rotation. (c) The net angular momentum of the system must
remain the same in spite of the inversion.
Lb
wh
(a) (b)
Lwh Lwh
ω
wh
ω
ω
Lb
Lwh Lwh
= +
(c)
Initial Final
b
The student now has
angular momentum,
and the net of these
two vectors equals
the initial vector.
that system to change about any vertical axis.
(Torques due to forces between the student and the
wheel as the student inverts the wheel are internal to the
system.) So, the system’s total angular momentum is con-
served about any vertical axis, including the rotation axis
through the stool.
L
:
tot
316 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Calculations: The conservation of is represented with
vectors in Fig. 11-20c. We can also write this conservation in
terms of components along a vertical axis as
Lb,f Lwh,f Lb,i Lwh,i, (11-35)
where iand frefer to the initial state (before inversion of
the wheel) and the final state (after inversion). Because
inversion of the wheel inverted the angular momentum
vector of the wheel’s rotation, we substitute Lwh,i for
Lwh,f. Then, if we set Lb,i 0 (because the student,
the stool, and the wheel’s center were initially at rest),
Eq. 11-35 yields
Lb,f 2Lwh,i.
L
:
tot
The rotational inertia of a disk rotating about its central
axis is given by Table 10-2cas . Substituting 6.00mfor
the mass M, our disk here has rotational inertia
. (11-36)
(We don’t have values for mand R, but we shall continue
with physics courage.)
From Eq. 10-33, we know that the rotational inertia of
the cockroach (a particle) is equal to mr2. Substituting the
cockroach’s initial radius ( ) and final radius
( ), we find that its initial rotational inertia about the
rotation axis is
(11-37)
and its final rotational inertia about the rotation axis is
. (11-38)
So, the cockroach–disk system initially has the rotational
inertia , (11-39)
and finally has the rotational inertia
. (11-40)
Next, we use Eq. 11-31 ( ) to write the fact that
the system’s final angular momentum Lfis equal to the sys-
tem’s initial angular momentum Li:
or .
After canceling the unknowns mand R, we come to
. (Answer)
Note that v decreased because part of the mass moved out-
ward, thus increasing that system’s rotational inertia.
vf1.37 rad/s
4.00mR2vf3.64mR2(1.50 rad/s)
IfvfIivi
LIv
IfIdIcf 4.00mR2
IiIdIci 3.64mR2
Icf mR2
Ici 0.64mR2
rR
r0.800R
Id3.00mR2
1
2MR2
Sample Problem 11.06 Conservation of angular momentum, cockroach on disk
In Fig.11-21, a cockroach with mass mrides on a disk of mass
6.00mand radius R. The disk rotates like a merry-go-round
around its central axis at angular speed The
cockroach is initially at radius , but then it crawls
out to the rim of the disk. Treat the cockroach as a particle.
What then is the angular speed?
KEY IDEAS
(1) The cockroach’s crawl changes the mass distribution (and
thus the rotational inertia) of the cockroach–disk system.
(2) The angular momentum of the system does not change
because there is no external torque to change it. (The forces
and torques due to the cockroach’s crawl are internal to the
system.) (3) The magnitude of the angular momentum of a
rigid body or a particle is given by Eq. 11-31 ( ).
Calculations: We want to find the final angular speed. Our
key is to equate the final angular momentum Lfto the initial
angular momentum Li, because both involve angular speed.
They also involve rotational inertia I.So, let’s start by finding
the rotational inertia of the system of cockroach and disk
before and after the crawl.
LIv
r0.800R
vi1.50 rad/s.
Additional examples, video, and practice available at WileyPLUS
Figure 11-21 A cockroach rides at radius ron a disk rotating like a
merry-go-round.
Rotation axis
Rr
i
ω
Using Eq. 11-31, we next substitute Ibvbfor Lb,f and Iwhvwh
for Lwh,i and solve for vb, finding
(Answer)
This positive result tells us that the student rotates counter-
clockwise about the stool axis as seen from overhead. If the
student wishes to stop rotating, he has only to invert the
wheel once more.
(2)(1.2 kgm2)(3.9 rev/s)
6.8 kgm21.4 rev/s.
b2Iwh
Ib
vwh
317
11-9 PRECESSION OF A GYROSCOPE
Precession of a Gyroscope
A simple gyroscope consists of a wheel fixed to a shaft and free to spin about the
axis of the shaft. If one end of the shaft of a nonspinning gyroscope is placed on a
support as in Fig. 11-22aand the gyroscope is released, the gyroscope falls by rotat-
ing downward about the tip of the support. Since the fall involves rotation, it is gov-
erned by Newton’s second law in angular form,which is given by Eq. 11-29:
(11-41)
This equation tells us that the torque causing the downward rotation (the fall)
changes the angular momentum of the gyroscope from its initial value of zero.
The torque is due to the gravitational force acting at the gyroscope’s center
of mass, which we take to be at the center of the wheel.The moment arm relative to
the support tip,located at Oin Fig. 11-22a,is .The magnitude of is
tMgr sin 90Mgr (11-42)
(because the angle between and is 90), and its direction is as shown in
Fig. 11-22a.
A rapidly spinning gyroscope behaves differently. Assume it is released with
the shaft angled slightly upward. It first rotates slightly downward but then, while
it is still spinning about its shaft, it begins to rotate horizontally about a vertical
axis through support point Oin a motion called precession.
Why Not Just Fall Over? Why does the spinning gyroscope stay aloft instead
of falling over like the nonspinning gyroscope? The clue is that when the spinning
gyroscope is released, the torque due to must change not an initial angular mo-
mentum of zero but rather some already existing nonzero angular momentum due
to the spin.
To see how this nonzero initial angular momentum leads to precession, we first
consider the angular momentum of the gyroscope due to its spin. To simplify the
situation,we assume the spin rate is so rapid that the angular momentum due to pre-
cession is negligible relative to . We also assume the shaft is horizontal when pre-
cession begins, as in Fig.11-22b.The magnitude of is given by Eq. 11-31:
LIv, (11-43)
where Iis the rotational moment of the gyroscope about its shaft and vis the an-
gular speed at which the wheel spins about the shaft. The vector points along
the shaft, as in Fig. 11-22b. Since is parallel to torque must be
perpendicular to .L
:t
:
r
:
,L
:L
:
L
:
L
:
L
:
Mg
:
r
:
Mg
:
t
:
r
:
Mg
:
:
L
:
t
:dL
:
dt .
Figure 11-22 (a) A nonspinning gyroscope
falls by rotating in an xz plane because of
torque . (b) A rapidly spinning gyroscope,
with angular momentum precesses
around the zaxis.Its precessional motion is
in the xy plane. (c) The change in
angular momentum leads to a rotation of
about O.
L
:
dL
:/dt
L
:,
t
:
x
y
z
τ
O
x
y
z
O
L
ω
=dL
___
dt
(a)
(b)
x
y
z
O
L(t)
dL
___
dt
d
φ
L(t+ dt)
Circular path
taken by head
of Lvector
(c)
τ
r
r
Mg
Mg
Support
11-9 PRECESSION OF A GYROSCOPE
After reading this module, you should be able to . . .
11.26 Identify that the gravitational force acting on a spinning
gyroscope causes the spin angular momentum vector (and
thus the gyroscope) to rotate about the vertical axis in a
motion called precession.
11.27 Calculate the precession rate of a gyroscope.
11.28 Identify that a gyroscope’s precession rate is
independent of the gyroscope’s mass.
A spinning gyroscope can precess about a vertical axis through its support at the rate
where Mis the gyroscope’s mass, ris the moment arm, Iis the rotational inertia, and vis the spin rate.
 Mgr
Iv,
Learning Objectives
Key Idea
318 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
M, mass cancels from Eq. 11-46; thus is independent of the mass.
Equation 11-46 also applies if the shaft of a spinning gyroscope is at an angle
to the horizontal. It holds as well for a spinning top, which is essentially a spinning
gyroscope at an angle to the horizontal.
According to Eq. 11-41, torque causes an incremental change in the
angular momentum of the gyroscope in an incremental time interval dt; that is,
(11-44)
However, for a rapidly spinning gyroscope, the magnitude of is fixed by
Eq. 11-43. Thus the torque can change only the direction of not its magnitude.
From Eq. 11-44 we see that the direction of is in the direction of per-
pendicular to . The only way that can be changed in the direction of
without the magnitude Lbeing changed is for to rotate around the zaxis as
shown in Fig. 11-22c. maintains its magnitude, the head of the vector follows
a circular path, and is always tangent to that path. Since must always
point along the shaft, the shaft must rotate about the zaxis in the direction of .
Thus we have precession. Because the spinning gyroscope must obey Newton’s
law in angular form in response to any change in its initial angular momentum, it
must precess instead of merely toppling over.
Precession. We can find the precession rate by first using Eqs. 11-44 and
11-42 to get the magnitude of :
dL tdt Mgr dt. (11-45)
As changes by an incremental amount in an incremental time interval dt, the shaft
and precess around the zaxis through incremental angle df. (In Fig. 11-22c, angle
dfis exaggerated for clarity.) With the aid of Eqs. 11-43 and 11-45, we find that dfis
given by
Dividing this expression by dt and setting the rate df/dt, we obtain
(precession rate). (11-46)
This result is valid under the assumption that the spin rate vis rapid. Note that
decreases as vis increased. Note also that there would be no precession if the
gravitational force did not act on the gyroscope, but because Iis a function ofMg
:
 Mgr
Iv
dfdL
LMgr dt
Iv.
L
:
L
:
dL
:
t
:
L
:
t
:
L
:
L
:L
:t
:
L
:
L
:t
:,dL
:L
:,
L
:
dL
:t
:dt.
dL
:
t
:
Rolling Bodies For a wheel of radius Rrolling smoothly,
vcom vR, (11-2)
where vcom is the linear speed of the wheel’s center of mass and vis
the angular speed of the wheel about its center. The wheel may
also be viewed as rotating instantaneously about the point Pof the
“road” that is in contact with the wheel. The angular speed of the
wheel about this point is the same as the angular speed of
the wheel about its center.The rolling wheel has kinetic energy
(11-5)
where Icom is the rotational inertia of the wheel about its center of
mass and Mis the mass of the wheel. If the wheel is being accelerated
but is still rolling smoothly,the acceleration of the center of mass
is related to the angular acceleration aabout the center with
acom aR. (11-6)
a
:
com
K1
2Icomv21
2
v2
com,
Review & Summary
If the wheel rolls smoothly down a ramp of angle u, its acceleration
along an xaxis extending up the ramp is
(11-10)
Torque as a Vector In three dimensions, torque is a vector
quantity defined relative to a fixed point (usually an origin); it is
(11-14)
where is a force applied to a particle and is a position vector lo-
cating the particle relative to the fixed point.The magnitude of is
(11-15, 11-16, 11-17)
where fis the angle between and is the component of
perpendicular to and is the moment arm of . The direction
of is given by the right-hand rule.t
:
F
:
r
r
:
,
F
:
F
r
:
,F
:
trF sin frF
rF,
t
:
r
:
F
:
t
:r
:F
:,
t
:
acom, x g sin u
1Icom /MR2.
319
QUESTIONS
Angular Momentum of a Particle The angular momentum
of a particle with linear momentum mass m, and linear velocity is
a vector quantity defined relative to a fixed point (usually an origin) as
(11-18)
The magnitude of is given by
(11-19)
(11-20)
(11-21)
where fis the angle between and and are the compo-
nents of and perpendicular to and is the perpendicular
distance between the fixed point and the extension of .The direc-
tion of is given by the right-hand rule for cross products.
Newton’s Second Law in Angular Form Newton’s second
law for a particle can be written in angular form as
(11-23)
where is the net torque acting on the particle and is the angu-
lar momentum of the particle.
Angular Momentum of a System of Particles The angu-
lar momentum of a system of particles is the vector sum of the
angular momenta of the individual particles:
(11-26)L
:
:
1
:
2 
:
n
n
i1
:
i.
L
:
:
t
:
net
t
:
net d
:
dt ,
:p
:
r
r
:
,v
:
p
:
v
p
p
:
,r
:
rprmv,
rprmv
rmv sin f
:
:r
:p
:m(r
:v
:
).
v
:
p
:
,
:The time rate of change of this angular momentum is equal to the
net external torque on the system (the vector sum of the torques
due to interactions with particles external to the system):
(system of particles). (11-29)
Angular Momentum of a Rigid Body For a rigid body
rotating about a fixed axis, the component of its angular
momentum parallel to the rotation axis is
LIv(rigid body, fixed axis). (11-31)
Conservation of Angular Momentum The angular mo-
mentum of a system remains constant if the net external torque
acting on the system is zero:
(isolated system) (11-32)
or (isolated system). (11-33)
This is the law of conservation of angular momentum.
Precession of a Gyroscope A spinning gyroscope can pre-
cess about a vertical axis through its support at the rate
(11-46)
where Mis the gyroscope’s mass, ris the moment arm, Iis the rota-
tional inertia, and vis the spin rate.
 Mgr
Iv,
L
:
iL
:
f
L
:a constant
L
:
t
:
net dL
:
dt
1Figure 11-23 shows three particles
of the same mass and the same constant
speed moving as indicated by the veloc-
ity vectors. Points a,b,c, and dform a
square, with point eat the center. Rank
the points according to the magnitude
of the net angular momentum of the
three-particle system when measured
about the points,greatest first.
2Figure 11-24 shows two parti-
cles Aand Bat xyz coordinates
(1 m, 1 m, 0) and (1 m, 0, 1 m).
Acting on each particle are three
numbered forces, all of the same
magnitude and each directed paral-
lel to an axis. (a) Which of the
forces produce a torque about the
origin that is directed parallel to y?
(b) Rank the forces according to
the magnitudes of the torques they
produce on the particles about the ori-
gin, greatest first.
3What happens to the initially sta-
tionary yo-yo in Fig. 11-25 if you pull it
via its string with (a) force (the line
of action passes through the point of
contact on the table, as indicated),
(b) force (the line of action passesF
:
1
F
:
2
above the point of contact), and (c) force (the line of action
passes to the right of the point of contact)?
4The position vector of a particle relative to a certain point
has a magnitude of 3 m, and the force on the particle has a mag-
nitude of 4 N. What is the angle between the directions of and
if the magnitude of the associated torque equals (a) zero and (b) 12
Nm?
5In Fig. 11-26, three forces of the
same magnitude are applied to a par-
ticle at the origin ( acts directly into
the plane of the figure). Rank the
forces according to the magnitudes of
the torques they create about (a)
point P1,(b) point P2,and (c) point P3,
greatest first.
6The angular momenta of a
particle in four situations are (1)
; (2) ; (3) ; (4) . In which situation
is the net torque on the particle (a) zero, (b) positive and con-
stant, (c) negative and increasing in magnitude (t0), and (d)
negative and decreasing in magnitude (t0)?
7A rhinoceros beetle rides the rim of a horizontal disk rotating
counterclockwise like a merry-go-round. If the beetle then walks
along the rim in the direction of the rotation, will the magnitudes
of the following quantities (each measured about the rotation axis)
increase, decrease, or remain the same (the disk is still rotating in
the counterclockwise direction): (a) the angular momentum of the
4/t26t2
3t4
(t)
F
:
1
F
:
r
:
F
:
r
:
F
:
3
Figure 11-23 Question 1.
c
d
e
ab
Figure 11-24 Question 2.
y
x
z
A
B6
5
42
1
3
F3
F2
F1
Figure 11-25 Question 3.
Questions
Figure 11-26 Question 5.
P1
P3
P2
F1
F3
F2
x
y
320 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
ing to the magnitude of the angular momentum of the particle
measured about them, greatest first.
beetledisk system, (b) the angular momentum and angular veloc-
ity of the beetle, and (c) the angular momentum and angular velocity
of the disk? (d) What are your answers if the beetle walks in the di-
rection opposite the rotation?
8Figure 11-27 shows an overhead
view of a rectangular slab that can
spin like a merry-go-round about its
center at O. Also shown are seven
paths along which wads of bubble
gum can be thrown (all with the
same speed and mass) to stick onto
the stationary slab. (a) Rank the paths according to the angular
speed that the slab (and gum) will have after the gum sticks, great-
est first. (b) For which paths will the angular momentum of the slab
(and gum) about Obe negative from
the view of Fig.11-27?
9Figure 11-28 gives the angular mo-
mentum magnitude Lof a wheel ver-
sus time t. Rank the four lettered time
intervals according to the magnitude
of the torque acting on the wheel,
greatest first.
10 Figure 11-29 shows a particle moving at constant velocity
and five points with their xy coordinates. Rank the points accord-
v
:
11 A cannonball and a marble roll smoothly from rest down an
incline. Is the cannonball’s (a) time to the bottom and (b) transla-
tional kinetic energy at the bottom more than, less than, or the
same as the marble’s?
12 A solid brass cylinder and a solid wood cylinder have the
same radius and mass (the wood cylinder is longer). Released to-
gether from rest,they roll down an incline.(a) Which cylinder reaches
the bottom first, or do they tie? (b) The wood cylinder is then short-
ened to match the length of the brass cylinder, and the brass cylinder
is drilled out along its long (central) axis to match the mass of the
wood cylinder.Which cylinder now wins the race, or do they tie?
(–3, 1) (9, 1)
x
(1, 3) c
a
bd
e
(–1, –2)
(4, –1)
y
v
A B C D
L
t
Figure 11-28 Question 9.
Figure 11-29 Question 10.
•5 A 1000 kg car has four 10 kg wheels. When the car is mov-
ing, what fraction of its total kinetic energy is due to rotation of the
wheels about their axles? Assume
that the wheels are uniform disks
of the same mass and size.Why do
you not need to know the radius
of the wheels?
••6 Figure 11-30 gives the speed
vversus time tfor a 0.500 kg ob-
ject of radius 6.00 cm that rolls
smoothly down a 30ramp. The
scale on the velocity axis is set by
vs 4.0 m/s. What is the rota-
tional inertia of the object?
••7 In Fig. 11-31, a solid cylin-
der of radius 10 cm and mass 12 kg
starts from rest and rolls without
slipping a distance L6.0 m down
a roof that is inclined at angle u
30. (a) What is the angular speed of
the cylinder about its center as it
leaves the roof? (b) The roof’s edge
is at height H5.0 m. How far hori-
zontally from the roof’s edge does
the cylinder hit the level ground?
ILW
ILW
Tutoring problem available (at instructor’s discretion) in WileyPLUS and WebAssign
SSM Worked-out solution available in Student Solutions Manual
••• Number of dots indicates level of problem difficulty
Additional information available in The Flying Circus of Physics and at flyingcircusofphysics.com
WWW Worked-out solution is at
ILW Interactive solution is at http://www.wiley.com/college/halliday
Problems
Module 11-1 Rolling as Translation and Rotation Combined
•1 A car travels at 80 km/h on a level road in the positive direction
of an xaxis. Each tire has a diameter of 66 cm. Relative to a woman
riding in the car and in unit-vector notation, what are the velocity
at the (a) center, (b) top, and (c) bottom of the tire and the magni-
tude aof the acceleration at the (d) center, (e) top, and (f) bottom
of each tire? Relative to a hitchhiker sitting next to the road and in
unit-vector notation, what are the velocity at the (g) center,
(h) top, and (i) bottom of the tire and the magnitude aof the
acceleration at the (j) center,(k) top, and (l) bottom of each tire?
•2 An automobile traveling at 80.0 km/h has tires of 75.0 cm di-
ameter.(a) What is the angular speed of the tires about their axles?
(b) If the car is brought to a stop uniformly in 30.0 complete turns
of the tires (without skidding), what is the magnitude of the angu-
lar acceleration of the wheels? (c) How far does the car move dur-
ing the braking?
Module 11-2 Forces and Kinetic Energy of Rolling
•3 A 140 kg hoop rolls along a horizontal floor so that the
hoop’s center of mass has a speed of 0.150 m/s. How much work
must be done on the hoop to stop it?
•4 A uniform solid sphere rolls down an incline. (a) What must be
the incline angle if the linear acceleration of the center of the
sphere is to have a magnitude of 0.10g? (b) If a frictionless block
were to slide down the incline at that angle, would its acceleration
magnitude be more than, less than, or equal to 0.10g? Why?
SSM
v
:
v
:
v (m/s)
vs
00.2 0.4
t (s)
0.6 0.8 1
Figure 11-30 Problem 6.
Figure 11-31 Problem 7.
L
H
θ
Figure 11-27 Question 8.
O
1
2
3
4 5 6
7
321
PROBLEMS
bowling ball of radius R11 cm
along a lane. The ball (Fig. 11-38)
slides on the lane with initial speed
vcom,0 8.5 m/s and initial angular
speed v00. The coefficient of ki-
netic friction between the ball and the lane is 0.21.The kinetic fric-
tional force acting on the ball causes a linear acceleration of the
ball while producing a torque that causes an angular acceleration
of the ball. When speed vcom has decreased enough and angular
speed vhas increased enough, the ball stops sliding and then rolls
smoothly. (a) What then is vcom in terms of v? During the sliding,
what are the ball’s (b) linear acceleration and (c) angular accelera-
tion? (d) How long does the ball slide? (e) How far does the ball
slide? (f) What is the linear speed of the ball when smooth rolling
begins?
•••16 Nonuniform cylindrical object. In Fig. 11-39, a cylindrical
object of mass Mand radius Rrolls smoothly from rest down a
ramp and onto a horizontal section. From there it rolls off the ramp
and onto the floor, landing a horizontal distance d0.506 m from
the end of the ramp. The initial height of the object is H0.90 m;
the end of the ramp is at height h0.10 m. The object consists of
an outer cylindrical shell (of a certain uniform density) that is
glued to a central cylinder (of a different uniform density).The ro-
tational inertia of the object can be expressed in the general form
IbMR2, but bis not 0.5 as it is for a cylinder of uniform density.
Determine b.
f
:
k
••8 Figure 11-32 shows the po-
tential energy U(x) of a solid
ball that can roll along an xaxis.
The scale on the Uaxis is set by
Us 100 J. The ball is uniform,
rolls smoothly, and has a mass of
0.400 kg. It is released at x7.0 m
headed in the negative direction
of the xaxis with a mechanical
energy of 75 J. (a) If the ball can
reach x0 m, what is its speed
there, and if it cannot, what is its
turning point? Suppose, instead, it is headed in the positive direc-
tion of the xaxis when it is released at x7.0 m with 75 J. (b) If
the ball can reach x13 m, what is
its speed there, and if it cannot, what
is its turning point?
••9 In Fig. 11-33, a solid ball
rolls smoothly from rest (starting at
height H6.0 m) until it leaves the
horizontal section at the end of the
track, at height h2.0 m. How far
horizontally from point Adoes the
ball hit the floor?
••10 A hollow sphere of radius 0.15 m, with rotational inertia
I0.040 kgm2about a line through its center of mass, rolls
without slipping up a surface inclined at 30to the horizontal. At
a certain initial position, the sphere’s total kinetic energy is 20 J.
(a) How much of this initial kinetic energy is rotational?
(b) What is the speed of the center of mass of the sphere at the
initial position? When the sphere has moved 1.0 m up the incline
from its initial position, what are (c) its total kinetic energy and
(d) the speed of its center of mass?
••11 In Fig. 11-34, a constant hori-
zontal force of magnitude 10 N is
applied to a wheel of mass 10 kg and
radius 0.30 m. The wheel rolls
smoothly on the horizontal surface,
and the acceleration of its center of
mass has magnitude 0.60 m/s2. (a) In
unit-vector notation, what is the fric-
tional force on the wheel? (b) What is the rotational inertia of the
wheel about the rotation axis through its center of mass?
F
:
app
rolls smoothly from rest down a ramp and onto a circular loop of
radius 0.48 m. The initial height of the ball is h0.36 m. At the
loop bottom, the magnitude of the normal force on the ball is
2.00Mg. The ball consists of an outer spherical shell (of a certain
uniform density) that is glued to a central sphere (of a different
uniform density). The rotational inertia of the ball can be ex-
pressed in the general form IbMR2, but bis not 0.4 as it is for a
ball of uniform density. Determine b.
•••14 In Fig. 11-37, a small, solid, uniform ball is to be shot
from point Pso that it rolls smoothly along a horizontal path, up
along a ramp, and onto a plateau.Then it leaves the plateau hori-
zontally to land on a game board, at a horizontal distance dfrom
the right edge of the plateau. The vertical heights are h15.00
cm and h21.60 cm. With what speed must the ball be shot at
point Pfor it to land at d6.00 cm?
Figure 11-32 Problem 8.
Figure 11-33 Problem 9.
H
h
A
02 4 6 8 101214
Us
U (J)
x (m)
Figure 11-37 Problem 14.
Figure 11-36 Problem 13.
Figure 11-35 Problem 12.
hRQ
h
Ball
P
h1d
h2
Figure 11-34 Problem 11.
Fapp
x
•••15 A bowler throws a
Figure 11-39 Problem 16.
Hh
d
x
fkvcom
Figure 11-38 Problem 15.
••12 In Fig. 11-35, a solid brass
ball of mass 0.280 g will roll
smoothly along a loop-the-loop
track when released from rest along
the straight section. The circular
loop has radius R14.0 cm, and the
ball has radius rR. (a) What is hif
the ball is on the verge of leaving
the track when it reaches the top of
the loop? If the ball is released at
height h6.00R, what are the (b)
magnitude and (c) direction of the
horizontal force component acting
on the ball at point Q?
•••13 Nonuniform ball. In Fig. 11-
36, a ball of mass Mand radius R
322 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
position vector .What
are (a) the torque on the particle about
the origin, in unit-vector notation, and (b)
the angle between the directions of
and ?
Module 11-5 Angular Momentum
•26 At the instant of Fig. 11-40, a 2.0 kg
particle Phas a position vector of magni-
tude 3.0 m and angle u145and a velocity
vector of magnitude 4.0 m/s and angle
u230. Force of magnitude 2.0 N andF
:
,
v
:
r
:
F
:r
:
r
:(3.0 m)i
ˆ(4.0 m)j
ˆ
(b) force with components F2x0,F2y2.0 N,F2z4.0 N?
••22 A particle moves through an xyz coordinate system while
a force acts on the particle. When the particle has the position
vector the force is given
by and the corresponding torque
about the origin is
Determine Fx.
••23 Force acts on a pebble with position
vector relative to the origin. In unit-vec-
tor notation, what is the resulting torque on the pebble about (a)
the origin and (b) the point (2.0 m, 0, 3.0 m)?
••24 In unit-vector notation, what is the torque about the origin
on a jar of jalapeño peppers located at coordinates (3.0 m, 2.0 m,
4.0 m) due to (a) force (b)
force and (c) the vector
sum of and ? (d) Repeat part (c) for the torque about the
point with coordinates (3.0 m, 2.0 m, 4.0 m).
••25 Force acts on a particle withF
:(8.0 N)i
ˆ(6.0 N)j
ˆ
SSM
F
:
2
F
:
1
(5.0 N)k
ˆ,F
:
2(3.0 N)i
ˆ(4.0 N)j
ˆ
(4.0 N)j
ˆ(5.0 N)k
ˆ,F
:
1(3.0 N)i
ˆ
r
:(0.50 m)j
ˆ(2.0 m)k
ˆ
F
:(2.0 N)i
ˆ(3.0 N)k
ˆ
(2.00 Nm)j
ˆ(1.00 Nm)k
ˆ .
t
:(4.00 Nm)i
ˆ
F
:Fxi
ˆ(7.00 N)j
ˆ(6.00 N)k
ˆ
r
:(2.00 m)i
ˆ(3.00 m)j
ˆ(2.00 m)k
ˆ,
F
:
2
Module 11-3 The Yo-Yo
•17 A yo-yo has a rotational inertia of 950 g cm2and
a mass of 120 g. Its axle radius is 3.2 mm, and its string is 120 cm
long. The yo-yo rolls from rest down to the end of the string.
(a) What is the magnitude of its linear acceleration? (b) How long
does it take to reach the end of the string? As it reaches the end of
the string, what are its (c) linear speed, (d) translational kinetic en-
ergy, (e) rotational kinetic energy, and (f) angular speed?
•18 In 1980, over San Francisco Bay, a large yo-yo was
released from a crane. The 116 kg yo-yo consisted of two uniform
disks of radius 32 cm connected by an axle of radius 3.2 cm. What
was the magnitude of the acceleration of the yo-yo during (a) its
fall and (b) its rise? (c) What was the tension in the cord on which
it rolled? (d) Was that tension near the cord’s limit of 52 kN?
Suppose you build a scaled-up version of the yo-yo (same shape
and materials but larger). (e) Will the magnitude of your yo-yo’s
acceleration as it falls be greater than, less than, or the same as that
of the San Francisco yo-yo? (f) How about the tension in the cord?
Module 11-4 Torque Revisited
•19 In unit-vector notation, what is the net torque about the ori-
gin on a flea located at coordinates (0, 4.0 m, 5.0 m) when forces
and act on the flea?
•20 A plum is located at coordinates (2.0 m, 0, 4.0 m). In unit-
vector notation, what is the torque about the origin on the plum if
that torque is due to a force whose only component is (a) Fx
6.0 N, (b) Fx6.0 N, (c) Fz6.0 N, and (d) Fz6.0 N?
•21 In unit-vector notation, what is the torque about the origin on
a particle located at coordinates (0, 4.0 m, 3.0 m) if that torque is
due to (a) force with components F1x2.0 N, F1yF1z0, andF
:
1
F
:
F
:
2(2.0 N)j
ˆ
F
:
1(3.0 N)k
ˆ
SSM
that has position vector and velocity vector
. About the origin and in unit-vector nota-
tion, what are (a) the object’s angular momentum and (b) the
torque acting on the object?
•28 A 2.0 kg particle-like object moves in a plane with velocity
components vx30 m/s and vy60 m/s as it passes through the
point with (x,y) coordinates of (3.0, 4.0) m. Just then, in unit-
vector notation, what is its angular momentum relative to (a) the
origin and (b) the point located at (2.0, 2.0) m?
•29 In the instant of Fig. 11-41,
two particles move in an xy plane.
Particle P1has mass 6.5 kg and
speed v12.2 m/s, and it is at dis-
tance d11.5 m from point O.
Particle P2has mass 3.1 kg and speed
v23.6 m/s, and it is at distance d2
2.8 m from point O. What are the
(a) magnitude and (b) direction of the
net angular momentum of the two particles about O?
••30 At the instant the displacement of a 2.00 kg object relative
to the origin is its veloc-
ity is and it is sub-
ject to a force Find (a) the
acceleration of the object, (b) the angular momentum of the object
about the origin, (c) the torque about the origin acting on the ob-
ject, and (d) the angle between the velocity of the object and the
force acting on the object.
••31 In Fig. 11-42, a 0.400 kg ball is
shot directly upward at initial speed 40.0
m/s. What is its angular momentum
about P, 2.00 m horizontally from the
launch point, when the ball is (a) at
maximum height and (b) halfway back
to the ground? What is the torque on the ball about Pdue to the
gravitational force when the ball is (c) at maximum height and
(d) halfway back to the ground?
Module 11-6 Newton’s Second Law in Angular Form
•32 A particle is acted on by two torques about the origin:
:
1
(4.00 N)k
ˆ.(8.00 N)j
ˆ
F
:(6.00 N)i
ˆ
(3.00 m/s)k
ˆ
v
:(6.00 m/s)i
ˆ(3.00 m/s)j
ˆ
d
:(2.00 m)i
ˆ(4.00 m)j
ˆ(3.00 m)k
ˆ,
ILW
v
:(5.0i
ˆ5.0k
ˆ) m/s
r
:(2.0i
ˆ2.0k
ˆ) m
Figure 11-40
Problem 26.
y
x
O
1
P
F
v
r
θ
2
θ
3
θ
Figure 11-41 Problem 29.
P2
Od2
d1v2
v1
P1
y
x
Ball
P
y
x
Figure 11-42 Problem 31.
has a magnitude of 2.0 Nm and is directed in the positive direc-
tion of the xaxis, and has a magnitude of 4.0 N m and is
directed in the negative direction of the yaxis. In unit-vector
notation, find where is the angular momentum of the
particle about the origin.
•33 At time t0, a 3.0 kg particle with velocity
ILWWWWSSM
:
d
:/dt,
t
:
2
angle u330, acts on P. All three vectors lie in the xy plane.
About the origin, what are the (a) magnitude and (b) direction of
the angular momentum of Pand the (c) magnitude and (d) direc-
tion of the torque acting on P?
•27 At one instant, force acts on a 0.25 kg objectF
:4.0j
ˆ N
SSM
is at x3.0 m, y8.0 m. It is pulled
by a 7.0 N force in the negative xdirection.About the origin, what are
(a) the particle’s angular momentum,(b) the torque acting on the par-
ticle,and (c) the rate at which the angular momentum is changing?
•34 A particle is to move in an xy plane, clockwise around the
origin as seen from the positive side of the zaxis.In unit-vector nota-
tion, what torque acts on the particle if the magnitude of its angular
momentum about the origin is (a) 4.0 kgm2/s, (b) 4.0t2kg m2/s,
(c) and (d) 4.0/t2kgm2/s?4.02t kgm2/s,
m/s)j
ˆ
v
:(5.0 m/s)i
ˆ(6.0
323
PROBLEMS
Module 11-8 Conservation of Angular Momentum
•43 In Fig. 11-47, two skaters, each
of mass 50 kg, approach each other
along parallel paths separated by
3.0 m. They have opposite velocities
of 1.4 m/s each. One skater carries
one end of a long pole of negligible
mass, and the other skater grabs the
other end as she passes. The skaters
then rotate around the center of the pole. Assume that the friction
between skates and ice is negligible. What are (a) the radius of the
circle, (b) the angular speed of the skaters, and (c) the kinetic energy
of the two-skater system? Next, the skaters pull along the pole until
they are separated by 1.0 m. What then are (d) their angular speed
and (e) the kinetic energy of the system? (f) What provided the en-
ergy for the increased kinetic energy?
•44 A Texas cockroach of mass 0.17 kg runs counterclockwise
around the rim of a lazy Susan (a circular disk mounted on a vertical
axle) that has radius 15 cm, rotational inertia 5.0 103kg m2, and
frictionless bearings. The cockroach’s speed (relative to the ground)
is 2.0 m/s,and the lazy Susan turns clockwise with angular speed v0
2.8 rad/s. The cockroach finds a bread crumb on the rim and, of
course, stops. (a) What is the angular speed of the lazy Susan after the
cockroach stops? (b) Is mechanical energy conserved as it stops?
•45 A man stands on a platform that is rotating (with-
out friction) with an angular speed of 1.2 rev/s; his arms are
outstretched and he holds a brick in each hand. The rotational iner-
tia of the system consisting of the man, bricks, and platform about
the central vertical axis of the platform is 6.0 kgm2.If by moving the
bricks the man decreases the rotational inertia of the system to 2.0
kgm2, what are (a) the resulting angular speed of the platform and
(b) the ratio of the new kinetic energy of the system to the original
kinetic energy? (c) What source provided the added kinetic energy?
•46 The rotational inertia of a collapsing spinning star drops to
its initial value. What is the ratio of the new rotational kinetic en-
ergy to the initial rotational kinetic energy?
1
3
WWWSSM
•37 In Fig. 11-44, three particles
of mass m23 g are fastened to
three rods of length d12 cm and
negligible mass. The rigid assembly
rotates around point Oat the angu-
lar speed v0.85 rad/s. About O,
what are (a) the rotational inertia
of the assembly, (b) the magnitude
of the angular momentum of the middle particle, and (c) the mag-
nitude of the angular momentum of the asssembly?
•38 A sanding disk with rotational inertia 1.2 103kg m2is
attached to an electric drill whose motor delivers a torque of mag-
nitude 16 Nm about the central axis of the disk. About that axis
and with the torque applied for 33 ms, what is the magnitude of the
(a) angular momentum and (b) angular velocity of the disk?
•39 The angular momentum of a flywheel having a rota-
tional inertia of 0.140 kg m2about its central axis decreases from
3.00 to 0.800 kgm2/s in 1.50 s. (a) What is the magnitude of the av-
erage torque acting on the flywheel about its central axis during
this period? (b) Assuming a constant angular acceleration, through
what angle does the flywheel turn? (c) How much work is done on
the wheel? (d) What is the average power of the flywheel?
••40 A disk with a rotational inertia of rotates like
a merry-go-round while undergoing a time-dependent torque
given by .At
time 1.00 s, its angular momen-
tum is . What is its an-
gular momentum at s?
••41 Figure 11-45 shows a rigid
structure consisting of a circular
hoop of radius Rand mass m, and a
square made of four thin bars, each
of length Rand mass m. The rigid
structure rotates at a constant speed
about a vertical axis,with a period of
t3.00
5.00 kgm2/s
t
(5.00 2.00t) Nmt
7.00 kgm2
SSM
••35 At time t, the vector gives the
position of a 3.0 kg particle relative to the origin of an xy coordinate
system ( is in meters and tis in seconds). (a) Find an expression for
the torque acting on the particle relative to the origin. (b) Is the
magnitude of the particle’s angular momentum relative to the origin
increasing,decreasing,or unchanging?
Module 11-7 Angular Momentum of a Rigid Body
•36 Figure 11-43 shows three rotating, uniform disks that are cou-
pled by belts.One belt runs around the rims of disks Aand C.Another
belt runs around a central hub on disk Aand the rim of disk B.The
belts move smoothly without slippage on the rims and hub.Disk Ahas
radius R; its hub has radius 0.5000R; disk Bhas radius 0.2500R; and
disk Chas radius 2.000R. Disks Band Chave the same density (mass
per unit volume) and thickness.What is the ratio of the magnitude of
the angular momentum of disk Cto that of disk B?
r
:
r
:4.0t2i
ˆ(2.0t6.0t2)j
ˆrotation of 2.5 s. Assuming R0.50 m and m2.0 kg, calculate
(a) the structure’s rotational inertia about the axis of rotation and
(b) its angular momentum about that axis.
••42 Figure 11-46 gives the torque tthat acts on an initially stationary
disk that can rotate about its center like a merry-go-round. The scale
on the taxis is set by ts4.0 Nm.What is the angular momentum of
the disk about the rotation axis at times (a) t7.0 s and (b) t20 s?
Belt
Belt
A
B
C
O
m
m
m
d
ω
dd
Rotation axis
R2R
Figure 11-43 Problem 36.
Figure 11-44 Problem 37.
Figure 11-45 Problem 41.
04 8 12 16 20
t (s)
s
τ
τ
(N m)
Figure 11-46 Problem 42.
Figure 11-47 Problem 43.
324 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
•47 A track is mounted on a
large wheel that is free to turn with
negligible friction about a vertical
axis (Fig. 11-48). A toy train of mass
mis placed on the track and, with the
system initially at rest, the train’s
electrical power is turned on. The train reaches speed 0.15 m/s with
respect to the track. What is the wheel’s angular speed if its mass is
1.1mand its radius is 0.43 m? (Treat it as a hoop, and neglect the
mass of the spokes and hub.)
•48 A Texas cockroach walks from
the center of a circular disk (that ro-
tates like a merry-go-round without
external torques) out to the edge at
radius R. The angular speed of the
cockroach–disk system for the walk is
given in Fig. 11-49 (va= 5.0 rad/s and
vb= 6.0 rad/s). After reaching R,
what fraction of the rotational inertia
of the disk does the cockroach have?
•49 Two disks are mounted (like a merry-go-round) on low-
friction bearings on the same axle and can be brought together so
that they couple and rotate as one unit. The first disk, with rota-
tional inertia 3.30 kg m2about its central axis, is set spinning coun-
terclockwise at 450 rev/min. The second disk, with rotational inertia
6.60 kgm2about its central axis, is set spinning counterclockwise
at 900 rev/min.They then couple together. (a) What is their angular
speed after coupling? If instead the second disk is set spinning
clockwise at 900 rev/min, what are their (b) angular speed and
(c) direction of rotation after they couple together?
•50 The rotor of an electric motor has rotational inertia Im
2.0 103kg m2about its central axis. The motor is used to
change the orientation of the space probe in which it is mounted.
The motor axis is mounted along the central axis of the probe; the
probe has rotational inertia Ip12 kgm2about this axis.
Calculate the number of revolutions of the rotor required to turn
the probe through 30about its central axis.
•51 A wheel is rotating freely at angular speed
800 rev/min on a shaft whose rotational inertia is negligible. A sec-
ond wheel, initially at rest and with twice the rotational inertia of the
first, is suddenly coupled to the same shaft. (a) What is the angular
speed of the resultant combination of the shaft and two wheels?
(b) What fraction of the original rotational kinetic energy is lost?
••52 A cockroach of mass mlies on the rim of a uniform disk of
mass 4.00mthat can rotate freely about its center like a merry-go-
round. Initially the cockroach and disk rotate together with an angu-
lar velocity of 0.260 rad/s. Then the cockroach walks halfway to the
center of the disk. (a) What then is the angular velocity of the cock-
roachdisk system? (b) What is the ratio K/K0of the new kinetic en-
ergy of the system to its initial kinetic energy? (c) What accounts for
the change in the kinetic energy?
••53 In Fig. 11-50 (an overhead
view), a uniform thin rod of length
0.500 m and mass 4.00 kg can rotate
in a horizontal plane about a verti-
cal axis through its center. The rod is
at rest when a 3.00 g bullet traveling
in the rotation plane is fired into one
end of the rod. In the view from
ILWSSM
SSM above, the bullet’s path makes angle u60.0with the rod (Fig. 11-
50). If the bullet lodges in the rod and the angular velocity of the rod
is 10 rad/s immediately after the collision, what is the bullet’s speed
just before impact?
••54 Figure 11-51 shows an
overhead view of a ring that can
rotate about its center like a merry-
go-round. Its outer radius R2is
0.800 m, its inner radius R1is R2/2.00,
its mass Mis 8.00 kg, and the mass of
the crossbars at its center is neg-
ligible. It initially rotates at an angu-
lar speed of 8.00 rad/s with a cat of
mass mM/4.00 on its outer edge, at
radius R2. By how much does the cat increase the kinetic energy of
the catring system if the cat crawls to the inner edge, at radius R1?
••55 A horizontal vinyl record of mass 0.10 kg and radius 0.10 m
rotates freely about a vertical axis through its center with an angu-
lar speed of 4.7 rad/s and a rotational inertia of 5.0 104kgm2.
Putty of mass 0.020 kg drops vertically onto the record from above
and sticks to the edge of the record. What is the angular speed of
the record immediately afterwards?
••56 In a long jump, an athlete leaves the ground with an
initial angular momentum that tends to rotate her body forward,
threatening to ruin her landing.To counter this tendency, she rotates
her outstretched arms to “take up” the angular momentum (Fig. 11-
18). In 0.700 s, one arm sweeps through 0.500 rev and the other arm
sweeps through 1.000 rev.Treat each arm as a thin rod of mass 4.0 kg
and length 0.60 m, rotating around one end. In the athlete’s reference
frame, what is the magnitude of the total angular momentum of the
arms around the common rotation axis through the shoulders?
••57 A uniform disk of mass 10mand radius 3.0rcan rotate
freely about its fixed center like a merry-go-round.A smaller uni-
form disk of mass mand radius rlies on top of the larger disk,
concentric with it. Initially the two disks rotate together with an an-
gular velocity of 20 rad/s.Then a slight disturbance causes the smaller
disk to slide outward across the larger disk, until the outer edge of the
smaller disk catches on the outer edge of the larger disk. Afterward,
the two disks again rotate together (without further sliding).(a) What
then is their angular velocity about the center of the larger disk? (b)
What is the ratio K/K0of the new kinetic energy of the two-disk sys-
tem to the system’s initial kinetic energy?
••58 A horizontal platform in the shape of a circular disk rotates on
a frictionless bearing about a vertical axle through the center of the
disk. The platform has a mass of 150 kg, a radius of 2.0 m, and a rota-
tional inertia of 300 kgm2about the axis of rotation.A 60 kg student
walks slowly from the rim of the platform toward the center.If the an-
gular speed of the system is 1.5 rad/s when the student starts at the
rim,what is the angular speed when she is 0.50 m from the center?
••59 Figure 11-52 is an overhead
view of a thin uniform rod of length
0.800 m and mass Mrotating horizon-
tally at angular speed 20.0 rad/s about
an axis through its center. A particle
of mass M/3.00 initially attached to
one end is ejected from the rod and travels along a path that is per-
pendicular to the rod at the instant of ejection. If the particle’s speed
vpis 6.00 m/s greater than the speed of the rod end just after ejec-
tion, what is the value of vp?
Figure 11-48 Problem 47.
ω
(rad/s)
ω
b
ω
a
0
Radial distance R
Figure 11-49 Problem 48.
Axis
θ
R1
R2
Figure 11-51 Problem 54.
Figure 11-50 Problem 53.
Rotation
axis
Figure 11-52 Problem 59.
325
PROBLEMS
••60 In Fig. 11-53, a 1.0 g bullet is fired
into a 0.50 kg block attached to the end
of a 0.60 m nonuniform rod of mass
0.50 kg. The blockrodbullet system
then rotates in the plane of the figure,
about a fixed axis at A. The rotational
inertia of the rod alone about that axis
at Ais 0.060 kgm2.Treat the block as a
particle. (a) What then is the rotational
inertia of the blockrodbullet system
about point A? (b) If the angular speed
of the system about Ajust after impact
is 4.5 rad/s, what is the bullet’s speed
just before impact?
••61 The uniform rod (length 0.60 m,
mass 1.0 kg) in Fig. 11-54 rotates in the
plane of the figure about an axis through
one end, with a rotational inertia of 0.12
kgm2. As the rod swings through its
lowest position, it collides with a 0.20 kg
putty wad that sticks to the end of the
rod.If the rod’s angular speed just before
collision is 2.4 rad/s, what is the angular
speed of the rodputty system immedi-
ately after collision?
•••62 During a jump to his partner, an aerialist is to
tangent to the outer edge of the merry-go-round, as shown. What is
the angular speed of the merry-go-round just after the ball is caught?
•••64 A ballerina begins a tour jeté (Fig. 11-19a) with angu-
lar speed and a rotational inertia consisting of two parts:
for her leg extended outward at angle
to her body and for the rest of her body (pri-
marily her trunk). Near her maximum height she holds both legs at
angle to her body and has angular speed (Fig. 11-19b).
Assuming that has not changed, what is the ratio ?
•••65 Two 2.00 kg balls are attached to the ends of a
thin rod of length 50.0 cm and negli-
gible mass.The rod is free to rotate in
a vertical plane without friction
about a horizontal axis through its
center.With the rod initially horizon-
tal (Fig. 11-57), a 50.0 g wad of wet
putty drops onto one of the balls, hit-
ting it with a speed of 3.00 m/s and then sticking to it. (a) What is the
angular speed of the system just after the putty wad hits? (b) What is
the ratio of the kinetic energy of the system after the collision to that
of the putty wad just before? (c) Through what angle will the system
rotate before it momentarily stops?
WWWSSM
vf/vi
Itrunk
vf
u30.0
Itrunk 0.660 kgm2
u90.0Ileg 1.44 kgm2
vi
A
Bullet
Block
Rod
Figure 11-53 Problem 60.
Figure 11-54 Problem 61.
Rod
Rotation axis
Release
Catch
Parabolic
path of
aerialist
Tuck
I1
I1
ω
1
I2
ω
2
ω
1
Figure 11-55 Problem 62.
Figure 11-57 Problem 65.
Putty wad
Rotation
axis
θ
O
h
Figure 11-58 Problem 66.
Figure 11-59 Problem 67.
•••63 In Fig. 11-56, a 30 kg child
stands on the edge of a stationary
merry-go-round of radius 2.0 m.
The rotational inertia of the merry-
go-round about its rotation axis is
150 kgm2. The child catches a ball
of mass 1.0 kg thrown by a friend.
Just before the ball is caught, it has a
horizontal velocity of magnitude
12 m/s, at angle f37with a line
v
:
Figure 11-56 Problem 63.
φ
Child
Ball
v
make a quadruple somersault lasting a time t1.87 s. For the first
and last quarter-revolution, he is in the extended orientation
shown in Fig. 11-55, with rotational inertia I119.9 kgm2around
his center of mass (the dot). During the rest of the flight he is in a
tight tuck, with rotational inertia I23.93 kgm2.What must be his
angular speed v2around his center of mass during the tuck?
•••66 In Fig. 11-58, a small 50 g
block slides down a frictionless sur-
face through height h20 cm and
then sticks to a uniform rod of mass
100 g and length 40 cm.The rod pivots
about point Othrough angle u
before momentarily stopping. Find u.
•••67 Figure 11-59 is an over-
head view of a thin uniform rod of
length 0.600 m and mass Mrotating
horizontally at 80.0 rad/s counter-
clockwise about an axis through its center. A particle of mass
M/3.00 and traveling horizontally at speed 40.0 m/s hits the rod
and sticks. The particle’s path is perpendicular to the rod at the
instant of the hit, at a distance dfrom the rod’s center. (a) At
what value of dare rod and particle stationary after the hit?
(b) In which direction do rod and particle rotate if dis greater
than this value?
Module 11-9 Precession of a Gyroscope
••68 A top spins at 30 rev/s about an axis that makes an angle of
30with the vertical.The mass of the top is 0.50 kg, its rotational in-
ertia about its central axis is 5.0 104kgm2, and its center of
mass is 4.0 cm from the pivot point. If the spin is clockwise from an
overhead view, what are the (a) precession rate and (b) direction of
the precession as viewed from overhead?
••69 A certain gyroscope consists of a uniform disk with a 50 cm
radius mounted at the center of an axle that is 11 cm long and of
negligible mass.The axle is horizontal and supported at one end. If
the spin rate is 1000 rev/min, what is the precession rate?
Rotation axis Particle
d
326 CHAPTER 11 ROLLING, TORQUE, AND ANGULAR MOMENTUM
Additional Problems
70 A uniform solid ball rolls smoothly along a floor, then up a
ramp inclined at 15.0. It momentarily stops when it has rolled
1.50 m along the ramp.What was its initial speed?
71 In Fig. 11-60, a constant
horizontal force of magnitude 12
N is applied to a uniform solid cylin-
der by fishing line wrapped around
the cylinder.The mass of the cylinder
is 10 kg, its radius is 0.10 m, and the
cylinder rolls smoothly on the hori-
zontal surface. (a) What is the mag-
nitude of the acceleration of the center of mass of the cylinder? (b)
What is the magnitude of the angular acceleration of the cylinder
about the center of mass? (c) In unit-vector notation, what is the
frictional force acting on the cylinder?
72 A thin-walled pipe rolls along the floor.What is the ratio of its
translational kinetic energy to its rotational kinetic energy about
the central axis parallel to its length?
73 A 3.0 kg toy car moves along an xaxis with a velocity
given by with tin seconds. For t0, what are
(a) the angular momentum of the car and (b) the torque on
the car, both calculated about the origin? What are (c) and (d)
about the point (2.0 m, 5.0 m, 0)? What are (e) and (f) about
the point (2.0 m, 5.0 m, 0)?
74 A wheel rotates clockwise about its central axis with an angu-
lar momentum of 600 kgm2/s. At time t0, a torque of magni-
tude 50 Nm is applied to the wheel to reverse the rotation. At
what time tis the angular speed zero?
75 In a playground, there is a small merry-go-round of
radius 1.20 m and mass 180 kg. Its radius of gyration (see Problem
79 of Chapter 10) is 91.0 cm.A child of mass 44.0 kg runs at a speed
of 3.00 m/s along a path that is tangent to the rim of the initially
stationary merry-go-round and then jumps on. Neglect friction be-
tween the bearings and the shaft of the merry-go-round. Calculate
(a) the rotational inertia of the merry-go-round about its axis of
rotation, (b) the magnitude of the angular momentum of the run-
ning child about the axis of rotation of the merry-go-round, and
(c) the angular speed of the merry-go-round and child after the
child has jumped onto the merry-go-round.
76 A uniform block of granite in the shape of a book has face di-
mensions of 20 cm and 15 cm and a thickness of 1.2 cm. The density
(mass per unit volume) of granite is 2.64 g/cm3. The block rotates
around an axis that is perpendicular to its face and halfway between
its center and a corner. Its angular momentum about that axis is
0.104 kgm2/s. What is its rotational kinetic energy about that axis?
77 Two particles, each of mass 2.90 104kg and speed
5.46 m/s, travel in opposite directions along parallel lines separated
by 4.20 cm. (a) What is the magnitude Lof the angular momentum
of the two-particle system around a point midway between the two
lines? (b) Is the value different for a different location of the
point? If the direction of either particle is reversed, what are the
answers for (c) part (a) and (d) part (b)?
78 A wheel of radius 0.250 m, moving initially at 43.0 m/s, rolls to
a stop in 225 m. Calculate the magnitudes of its (a) linear accelera-
tion and (b) angular acceleration. (c) Its rotational inertia is 0.155
kgm2about its central axis. Find the magnitude of the torque
about the central axis due to friction on the wheel.
SSM
SSM
t
:
L
:t
:
L
:t
:
L
:v
:2.0t3i
ˆ m/s,
SSM
F
:
app
SSM
79 Wheels Aand Bin Fig. 11-61 are
connected by a belt that does not slip.
The radius of Bis 3.00 times the radius
of A. What would be the ratio of the
rotational inertias IA/IBif the two
wheels had (a) the same angular mo-
mentum about their central axes and
(b) the same rotational kinetic energy?
80 A 2.50 kg particle that is moving horizontally over a floor
with velocity (3.00 m/s)ˆ
j undergoes a completely inelastic colli-
sion with a 4.00 kg particle that is moving horizontally over the
floor with velocity (4.50 m/s)i
ˆ. The collision occurs at xy coordi-
nates (0.500 m, 0.100 m). After the collision and in unit-vector
notation, what is the angular momentum of the stuck-together par-
ticles with respect to the origin?
81 A uniform wheel
of mass 10.0 kg and radius
0.400 m is mounted rigidly
on a massless axle through
its center (Fig. 11-62). The
radius of the axle is 0.200
m, and the rotational inertia
of the wheelaxle combi-
nation about its central axis
is 0.600 kgm2.The wheel is
initially at rest at the top of a surface that is inclined at angle u
30.0with the horizontal; the axle rests on the surface while the
wheel extends into a groove in the surface without touching the
surface. Once released, the axle rolls down along the surface
smoothly and without slipping.When the wheelaxle combination
has moved down the surface by 2.00 m, what are (a) its rotational
kinetic energy and (b) its translational kinetic energy?
82 A uniform rod rotates in a horizontal plane about a vertical axis
through one end.The rod is 6.00 m long,weighs 10.0 N, and rotates at
240 rev/min. Calculate (a) its rotational inertia about the axis of rota-
tion and (b) the magnitude of its angular momentum about that axis.
83 A solid sphere of weight 36.0 N rolls up an incline at an angle
of 30.0. At the bottom of the incline the center of mass of the
sphere has a translational speed of 4.90 m/s. (a) What is the kinetic
energy of the sphere at the bottom of the incline? (b) How far does
the sphere travel up along the incline? (c) Does the answer to
(b) depend on the sphere’s mass?
84 Suppose that the yo-yo in Problem 17, instead of rolling
from rest, is thrown so that its initial speed down the string is
1.3 m/s. (a) How long does the yo-yo take to reach the end of
the string? As it reaches the end of the string, what are its (b) total
kinetic energy, (c) linear speed, (d) translational kinetic energy,
(e) angular speed, and (f) rotational kinetic energy?
85 A girl of mass Mstands on the rim of a frictionless merry-
go-round of radius Rand rotational inertia Ithat is not moving. She
throws a rock of mass mhorizontally in a direction that is tangent to
the outer edge of the merry-go-round.The speed of the rock, relative
to the ground, is v. Afterward, what are (a) the angular speed of the
merry-go-round and (b) the linear speed of the girl?
86 A body of radius Rand mass mis rolling smoothly with speed
von a horizontal surface. It then rolls up a hill to a maximum
height h. (a) If h3v2/4g, what is the body’s rotational inertia
about the rotational axis through its center of mass? (b) What
might the body be?
SSM
Figure 11-60 Problem 71.
Fishing line
x
Fapp
Figure 11-62 Problem 81.
Groove
Axle
Wheel
θ
Figure 11-61 Problem 79.
B
A